标签:ems span lin .com com lse online char lan
[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=1059
[算法]
二分图最大匹配
时间复杂度 : O(N^3)
[代码]
#include<bits/stdc++.h> using namespace std; #define MAXN 210 struct edge { int to , nxt; } e[MAXN * MAXN]; int n , tot; int match[MAXN],head[MAXN]; bool visited[MAXN]; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘; x *= f; } inline void addedge(int u,int v) { tot++; e[tot] = (edge){v,head[u]}; head[u] = tot; } inline bool hungary(int u) { for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (!visited[v]) { visited[v] = true; if (!match[v] || hungary(match[v])) { match[v] = u; return true; } } } return false; } int main() { int T; read(T); while (T--) { read(n); tot = 0; for (int i = 1; i <= n; i++) head[i] = 0; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { int x; read(x); if (x == 1) addedge(i,j); } } memset(match,0,sizeof(match)); int t = 0; for (int i = 1; i <= n; i++) { memset(visited,false,sizeof(visited)); if (hungary(i)) t++; } if (t == n) printf("Yes\n"); else printf("No\n"); } return 0; }
标签:ems span lin .com com lse online char lan
原文地址:https://www.cnblogs.com/evenbao/p/9734413.html