#include<iostream> #include<algorithm> #include<cstring> #define N 25005 using namespace std; typedef long long ll; ll road[2501][2501]; int main() { int n, m, s, t; bool visit[2501]; ll best[2501] = { 0 }; for (int i = 0; i <= 2500; i++) for (int j = 0; j <= 2500; j++) road[i][j] = 1e18; for (int i = 0; i <= 2500; i++) best[i] = 1e18; cin >> n >> m >> s >> t; best[s] = 0; while (m--) { int a, b; ll c; cin >> a >> b >> c; if (c<road[a][b]) road[a][b] = road[b][a] = c; } while (1) { int ma = 1e18, target = -1; for (int i = 1; i <= n; i++) { if (!visit[i] && ma>best[i]) { ma = best[i]; target = i; } } if (target == -1)break; visit[target] = 1; for (int i = 1; i <= n; i++) { if (!visit[i]) { best[i] = min(best[i], best[target] + road[target][i]); } } } cout << best[t];//output return 0; }对这个代码进行降重

使用include <iostream>时一定要加using namespace std

#include <iostream> #include<iomanip> using namespace std;九九乘法表

vc++2008编译不了#include头文件

### vc++2008 编译不了 #include<iostream.h> 头文件 #### 背景介绍在使用 Visual C++ 2008（以下简称 VC++ 2008）进行 C++ 编程时，可能会遇到无法编译包含 #include <iostream.h> 的代码的问题。这个问题主要...

#include <iostream> #include <algorithm> #include <cstdio> #include <cmath> #include <vector> #include <map> #include <vector> #include <string> #include <cstring> #define fast ios::sync_with_stdio(false),cin.tie(0) using namespace std; typedef pair<int, int> PII; typedef long long LL; const int N = 2e5+10; int T; string name[10] = {"Captain", "Priest", "Cook", "Doctor", "Engineer", "Hunter", "Gunner", "Navigator" }; map<string, int> mp; int main() { string s; cin >> s; mp[s] = 1; int n; scanf("%d", &n); for(int i = 0; i < n; i ++ ) { int j = 0; string s, t; getline(cin, s); for(int i = 0; s[i] != ':'; i ++ ) t += s[i]; mp[t] = 1; } sort(name, name + 8); int f = 0; for(int i = 0; i < 8; i ++ ) { if(mp[name[i]] == 0) cout << name[i] << endl, f ++; } if(!f) cout << "Ready" << endl; return 0; }转化为C语言代码

1. 删除头文件#include <iostream>，#include <algorithm>，#include <cstdio>，#include <cmath>，#include <vector>，#include <map>，#include <vector>，#include <string>，#include <cstring>。 2. 删除using...

#include <cstdio> #include<iostream> #include<queue> #include<algorithm> #include<cstring> #include<iomanip> #define inf 0x3f3f3f3f3f3f3f3fLL using namespace std; typedef long long ll; const int N=1e6; int n,k; int a[N]; int quick(int l,int r,int k) { if(l==r) return a[l]; int i=l-1,j=r+1,x=a[l]; while(i<j) { while(a[i++]<x); while(a[j--]>x); if(i<j) swap(a[i],a[j]); } int s=j-l+1; if(k<=s) return quick(l,j,k); return quick(j+1,r,k-s); } int main() { cin>>n>>k; for(int i=0;i<n;i++) { cin>>a[i]; } cout<<quick(0,n-1,k); return 0; }上述代码为什么报错

代码开头只有一行 #include <cstdio>，缺少其他可能需要的头文件，如 iostream、queue、algorithm、cstring 等。可以添加以下代码： c++ #include<iostream> #include<queue> #include<algorithm>...

求一个点到另外一个点的最小距离 #define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<algorithm> #include<vector> #include<cstring> #include<cstdio> #include<cmath> #include<cstdlib> #include<sstream> #include<string> #include<map> using namespace std; typedef long long ll; int dx[4] = { 0, 0, 1, -1 }; int dy[4] = { 1, -1, 0, 0 }; #define fo(i,l,r) for (int i = l; i <= r; ++i) #define F 10005 int n, m, s, t,x,y,ww; ll D[F],w[F][F],vis[F]; int main() { cin >> n >> m >> s >> t; fo(i, 1, n) fo(j, 1, n) { if (i != j) w[i][j] = 0x3f3f3f3f3f; } fo(i, 0, n)//注意0的初始化 D[i] = 0x3f3f3f3f3f; D[s] = 0;//从源点到i的距离 fo(i, 1, m) { scanf("%d%d%d", &x, &y, &ww); w[x][y] = ww; w[y][x] = ww; } //dijkstra算法 int k = 0; fo(i, 1, n) { k = 0; fo(j, 1, n) if (!vis[j] && D[j] < D[k])//一开始找源点 k = j; vis[k] = 1; fo(j, 1, n) if (!vis[j]&&D[k] + w[k][j] < D[j]) { D[j] = D[k] + w[k][j]; } } cout << D[t] << endl; return 0; }不用任何容器来实现这个代码

#include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<cmath> #include<cstdlib> #include<sstream> #include<string> #include<map> using namespace std; typedef long long ll...

#include <iostream> #include <algorithm> #include <cstdio> #include <cmath> #include <vector> #include <map> #include <string> #include <cstring> #define fast ios::sync_with_stdio(false),cin.tie(0) using namespace std; typedef pair<int, int> PII; typedef long long LL; const int N = 1010; int T; int f1, s1, f2, s2; vector v; string x[8] = {"11111111", "10000001", "10111101", "10111101", "10111101", "10111101", "10000001", "11111111"}; int main() { int n, m; scanf("%d %d", &n, &m); string s[N]; for(int i = 0; i < n; i ++ ) { cin >> s[i]; } for(int i = 0; i <= n - 8; i ++ ) { for(int j = 0; j <= m - 8; j ++ ) { int k, cnt = 0; for(k = 0; k < 8; k ++ ) { //cout << k << " " << x[k] << " " << s[i+k].substr(j, 8) << endl; if( x[k] == s[i+k].substr(j, 8) ) cnt ++; } if(cnt == 8) v.push_back({i, j}); } } sort(v.begin(), v.end()); for(int i = 0; i < 3; i ++ ) printf("%d %d\n", v[i].first, v[i].second); return 0; }

1. 读入两个整数 n 和 m（n 和 m 分别表示一个矩阵的行数和列数）。 2. 读入一个字符串数组 s，长度为 n，每个元素为长度为 m 的字符串。 3. 定义一个字符串数组 x，长度为 8，每个元素为长度为 8 的字符串。 4. ...

把#include <iostream> #include <algorithm> #include <cstdio> #include <cmath> #include <vector> #include <map> #include <vector> #include <string> #include <cstring> #define fast ios::sync_with_stdio(false),cin.tie(0) using namespace std; typedef pair<int, int> PII; typedef long long LL; const int N = 2e5+10; int T, t; string s[3]; void init() { /* #### #### #### # # # # # # # # #/ // 第一行，中间行，和最后一行 // 构造成 #### #### #### for(int i = 0; i < 3; i ++ ) { if(i) s[0] += " "; s[0] += "###"; for(int j = 0; j < t; j ++ ) s[0] += "#"; } // 上半部分 // 构造成 # # # # s[1] += "# "; for(int i = 0; i < t; i ++ ) s[1] += " "; s[1] += "# #"; for(int i = 0; i < 2 t + 5; i ++ ) s[1] += " "; s[1] += "#"; // 下半部分 // 构造成 # # # # # s[2] += "# "; for(int i = 0; i < t; i ++ ) s[2] += " "; s[2] += "# # "; for(int i = 0; i < t; i ++ ) s[2] += " "; s[2] += "#"; for(int i = 0; i < t + 3; i ++ ) s[2] += " "; s[2] += "#"; } int main() { scanf("%d", &t); int n = t * 2 + 5; init(); for(int i = 1; i <= n; i ++ ) { if(i == 1 || i == n || i == (n + 1)/2) cout << s[0] << endl; else if(i < (n + 1) / 2) cout << s[1] << endl; else cout << s[2] << endl; } return 0; } 这段代码变成C语言

把#include <iostream>改为#include <stdio.h>，把using namespace std;删除，将cout改为printf即可： c #include <stdio.h> #include <algorithm> #include <cstdio> #include <cmath> #include <vector> #...

#include <iostream> #include <cstring> #include <algorithm> using namespace std; const int N = 1510; int n; int h[N], e[N], ne[N], idx; int f[N][2]; bool not_root[N]; void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ; } void dfs(int u) { f[u][0] = 0, f[u][1] = 1; //initialize for (int i = h[u]; ~i; i = ne[i]) { int j = e[i]; dfs(j); f[u][0] += f[j][1]; f[u][1] += min(f[j][0], f[j][1]); } } int main() { while (~scanf("%d", &n)) { memset(h, -1, sizeof h); idx = 0; memset(not_root, 0, sizeof not_root); for (int i = 0; i < n; i ++ ) { int a, b, siz; scanf("%d:(%d) ", &a, &siz); while (siz -- ) { scanf("%d", &b); add(a, b); not_root[b] = true; } } int root = 0; while (not_root[root]) root ++ ; dfs(root); printf("%d\n", min(f[root][0], f[root][1])); } return 0; } #include <iostream> #include <vector> #include <set> #include <cstring> #define int long long #define debug cout << "xxx" << '\n'; using namespace std; typedef pair<int,int> PII; const int N = 1510; int n; vector<int> e[N]; // f[x][0] 代表当前位置不放士兵，f[x][1]代表当前位置放士兵 vector<vector<int>> f(N, vector<int>(2, 0)); vector<bool> v(N, false); int ans = 1e18; void dp(int x) { f[x][0] = 0, f[x][1] = 1; for(int i = 0; i < e[x].size(); i ++) { int y = e[x][i]; if(y == x) continue; dp(y); f[x][0] += f[y][1]; // 父节点没放士兵，那么子节点必须要放士兵 f[x][1] += min(f[y][1], f[y][0]); //父节点放士兵，子节点随便放不放 } return; } signed main() { ios::sync_with_stdio(false); cin.tie(0);cout.tie(0); while(cin >> n) { for(int i = 0; i <= n - 1; i ++) e[i].clear(); for(int i = 0; i <= n - 1; i ++) v[i] = false; for(int i = 1; i <= n; i ++) { string s; cin >> s; int x = s[0] - '0'; int k = s[3] - '0'; for(int j = 1; j <= k; j ++) { int y; cin >> y; e[x].push_back(y); v[y] = 1; } } int root = 0; for(int i = 0; i < n - 1; i ++) { if(!v[i]) { root = i; break; } } dp(root); cout << min(f[root][0], f[root][1]) << '\n'; } return 0; }为什么第一段代码对了，第二段代码错了

<think>嗯，用户给了两段C++代码，都是解决同一个问题，但第一段正确，第二段错误。需要找出第二段哪里出错了。首先，我得先理解这个问题是做什么的。看起来像是树形DP的问题，可能是在一个树结构上放置最少的士兵，...

#include<iostream> #include<algorithm> #include<cstring> using namespace std; #define INF 1<<30 typedef pair<int, int> PII; PII m[60]; int dist[60][60], mark[60][60]; void path(int l, int r) { if (l == r) return; int k = mark[l][r]; path(l, k); path(k + 1, r); printf("%d ", k + 1); } int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) scanf("%d,%d", &m[i].first, &m[i].second), dist[i][i] = 0; for (int len = 2; len <= n; len++) for (int l = 0; l + len - 1 < n; l++) { int r = l + len - 1; dist[l][r] = INF; for (int k = l; k < r; k++) { int res = dist[l][k] + dist[k + 1][r] + m[l].first * m[k].second * m[r].second; if (res < dist[l][r]) { dist[l][r] = res; mark[l][r] = k; } } } path(0, n - 1); puts(""); printf("%d\n", dist[0][n - 1]); return 0; }

这是一个动态规划求解矩阵链乘法问题的代码。代码中使用了一个二维数组 dist 来保存当前区间 [l,r] 的最小矩阵乘法代价，以及一个二维数组 mark 来保存当前区间 [l,r] 的最优决策...时间复杂度为 $O(n^3)$。

#include <iostream> #include <cstring> #include <algorithm> #define MAX_CAPACITY 100 // 最大容量限制 typedef struct { char no[8]; // 图书编号 char name[20]; // 书名 float price; // 价格 } Book; typedef struct { Book* elem; // 存储空间基址 int length; // 当前长度 int capacity; // 总容量 } SqList; // 初始化顺序表 void InitList(SqList& L, int size) { L.elem = new Book[size]; L.length = 0; L.capacity = size; } // 输入图书信息 void InputBooks(SqList& L, int n) { if (n > L.capacity) { std::cout << "超出最大容量！" << std::endl; return; } for (int i = 0; i < n; i++) { std::cout << "输入第" << i+1 << "本图书信息（编号书名价格）: "; std::cin >> L.elem[i].no >> L.elem[i].name >> L.elem[i].price; L.length++; } } // 显示所有图书 void DisplayAll(SqList& L) { if (L.length == 0) { std::cout << "暂无图书信息！" << std::endl; return; } std::cout << "\n编号\t书名\t价格" << std::endl; for (int i = 0; i < L.length; i++) { std::cout << L.elem[i].no << "\t" << L.elem[i].name << "\t" << L.elem[i].price << std::endl; } } // 按书名查找 void SearchByName(SqList& L, char* name) { bool found = false; for (int i = 0; i < L.length; i++) { if (strcmp(L.elem[i].name, name) == 0) { std::cout << "找到图书：" << L.elem[i].no << "\t" << L.elem[i].name << "\t" << L.elem[i].price << std::endl; found = true; } } if (!found) std::cout << "未找到该图书！" << std::endl; } // 插入图书 void InsertBook(SqList& L, int pos, Book book) { if (pos < 1 || pos > L.length + 1) { std::cout << "无效位置！" << std::endl; return; } if (L.length >= L.capacity) { std::cout << "存储空间已满！" << std::endl; return; } for (int i = L.length; i >= pos; i--) { L.elem[i] = L.elem[i 把这个代码补充完善

#include <algorithm> #define MAX_CAPACITY 100 // 最大容量限制 typedef struct { char no[8]; // 图书编号 char name[20]; // 书名 float price; // 价格 } Book; typedef struct { Book* elem; // 存储...

#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <set> #include <map> #include <queue> using namespace std; typedef long long ll; const ll INF = (1ll << 62); #define DEBUG(x) std::cerr << #x << " = " << x << std::endl #define int ll inline ll read() { ll f = 1, x = 0;char ch; do {ch = getchar();if (ch == '-')f = -1;} while (ch > '9' || ch < '0'); do {x = x * 10 + ch - '0';ch = getchar();} while (ch >= '0' && ch <= '9'); return f * x; } const int MAX_N = 100000 + 50; const int MAX_M = 300000 + 50; const int MAX_F = 30 + 5; struct EDGE { int to, next, w; } edge[MAX_M << 1]; int head[MAX_N], cnt; void addedge (int u, int v, int w) { edge[++cnt].to = v; edge[cnt].w = w; edge[cnt].next = head[u]; head[u] = cnt; } int f[MAX_N][MAX_F], g[MAX_N][MAX_F], h[MAX_N][MAX_F], dep[MAX_N]; inline void ckmax (int &x, int y) { x = (x > y ? x : y); } inline void dfs (int u, int fa, int w) { dep[u] = dep[fa] + 1; f[u][0] = fa; g[u][0] = w; h[u][0] = -INF; for (int i = 1; i <= 20; i ++ ) { f[u][i] = f[f[u][i - 1]][i - 1]; g[u][i] = max (g[u][i - 1], g[f[u][i - 1]][i - 1]); h[u][i] = max (h[u][i - 1], h[f[u][i - 1]][i - 1]); if (g[u][i - 1] > g[f[u][i - 1]][i - 1]) h[u][i] = max (h[u][i], g[f[u][i - 1]][i - 1]); else if (g[u][i - 1] < g[f[u][i - 1]][i - 1]) h[u][i] = max (h[u][i], g[u][i - 1]); } for (int i = head[u]; i; i = edge[i].next ) { int v = edge[i].to, w = edge[i].w; if (v == fa) continue; dfs (v, u, w); } } inline int LCA (int x, int y) { if (dep[x] < dep[y]) swap (x, y); for (int i = 20; i >= 0; i -- ) { if (dep[f[x][i]] >= dep[y]) x = f[x][i]; } if (x == y) return x; for (int i = 20; i >= 0; i -- ) { if (f[x][i] != f[y][i]) { x = f[x][i]; y = f[y][i]; } } return f[x][0]; } int n, m, fa[MAX_N], res, sum; struct Node { int u, v, w; bool operator < (const Node & rhs ) const { return w < rhs.w; } } a[MAX_M << 1]; inline int find (int x) { return x == fa[x] ? x : fa[x] = find (fa[x]); } inline void merge (int x, int y) { x = find (x); y = find (y); if (x == y) return; fa[x] = y; } bool vis[MAX_M]; inline void kruskal () { n = read (); m = read (); for (int i = 1; i <= m; i ++ ) { a[i].u = read (); a[i].v = read (); a[i].w = read (); } sort (a + 1, a + m + 1); for (int i = 1; i <= n; i ++ ) fa[i] = i; res = 0; for (int i = 1; i <= m; i ++ ) { if (find (a[i].u) != find (a[i].v)) { vis[i] = 1; res ++ ; merge (a[i].u, a[i].v); sum += a[i].w; addedge (a[i].u, a[i].v, a[i].w); addedge (a[i].v, a[i].u, a[i].w); } if (res == n - 1) break; } res = 0; dfs (1, 0, 0); } inline int qmax (int u, int v, int maxx) { int ans = -INF; for (int i = 18; i >= 0; i -- ) { if (dep[f[u][i]] >= dep[v]) { if (maxx != g[u][i]) ans = max (ans, g[u][i]); else ans = max (ans, h[u][i]); u = f[u][i]; } } return ans; } inline void ckmin (int &x, int y) { x = (x < y ? x : y); } inline void calc () { int ans = INF; for (int i = 1; i <= m; i ++ ) { if (vis[i]) continue; int lca = LCA (a[i].u, a[i].v); int max_u = qmax (a[i].u, lca, a[i].w); int max_v = qmax (a[i].v, lca, a[i].w); ckmin (ans, sum - max (max_u, max_v) + a[i].w); } printf ("%lld\n", ans); } signed main() { kruskal (); calc (); return 0; }为什么qmax函数需要循环 i 从大到小

<think>好的，我需要解释为什么在qmax函数中循环变量i是从大到小遍历的。首先，我得仔细看看代码的结构，特别是涉及到LCA（最近公共祖先）和倍增法的部分。这段代码看起来是在处理最小生成树的次小生成树问题。...

#include <iostream> #include <cstring> // 图书信息结构体 typedef struct { char no[8]; // 图书编号 char name[20]; // 书名 float price; // 价格 } Book; // 顺序表结构 typedef struct { Book* elem; // 存储空间基址 int length; // 当前长度 } SqList; // 初始化顺序表 void InitList(SqList& L, int size) { L.elem = new Book[size]; L.length = 0; } // (1) 输入图书信息 void InputBooks(SqList& L, int n) { for (int i = 0; i < n; i++) { std::cout << "输入第" << i+1 << "本图书信息（编号书名价格）: "; std::cin >> L.elem[i].no >> L.elem[i].name >> L.elem[i].price; L.length++; } } // (2) 显示所有图书 void DisplayAll(SqList& L) { for (int i = 0; i < L.length; i++) { std::cout << L.elem[i].no << "\t" << L.elem[i].name << "\t" << L.elem[i].price << std::endl; } } // (3) 按书名查找 void SearchByName(SqList& L, char* name) { for (int i = 0; i < L.length; i++) { if (strcmp(L.elem[i].name, name) == 0) { std::cout << "找到图书：" << L.elem[i].no << " 价格：" << L.elem[i].price << std::endl; return; } } std::cout << "未找到该图书！" << std::endl; }完善成完整的带代码

#include <algorithm> // 用于swap函数 void SortByPrice(SqList &L, bool ascending = true) { for (int i = 0; i < L.length - 1; i++) { for (int j = 0; j < L.length - i - 1; j++) { bool condition = ...

优化以下代码#include<iostream> using namespace std; #include<algorithm> #define MAXSIZE 100 int father[MAXSIZE];//存放父亲节点 typedef struct Mgraph { int s[MAXSIZE][MAXSIZE]; int n;//节点数 }; struct graph { int a, b, c; }arr[MAXSIZE]; bool compare(graph x, graph y) {//对边按照权值从小到大排序 return x.c < y.c; } int Find(int x) {//找到一个节点的根节点，并将其下面的所有节点的father都改为根节点。 while (x != father[x]) { int temp = x; x = father[x]; father[temp] = x; } return x; } int main() { Mgraph p; int m; cin >> p.n >> m; for (int i = 0; i < m; i++) { cin >> arr[i].a >> arr[i].b >> arr[i].c;//a,b间，需要c天 } for (int i = 1; i < p.n + 1; i++)//初始化father数组，将每个节点的父亲节点设置为自己 father[i] = i; sort(arr, arr + m, compare);//对边按照权值从小到大排序 for (int i = 0; i < m; i++) { if (Find(arr[i].a) != Find(arr[i].b)) father[Find(arr[i].a)] = Find(arr[i].b);//遍历每条边，如果该边的两个节点不在同一个连通块中，则将它们连通，并将它们的父亲节点设置为同一个节点。 if (Find(1) == Find(p.n)) {//如果1号节点和n号节点在同一个连通块中，则输出当前边的权值，并结束程序 cout << arr[i].c << endl; break; } } }

using namespace std; #define MAXSIZE 100010 struct Edge { int u, v, w; bool operator < (const Edge& e) const { return w < e.w; } }; int father[MAXSIZE]; int n, m; int find(int x) { if (x != ...

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