摘要：
#pragma comment(linker, "/STACK:102400000,102400000")#include#include#include#includeusing namespace std;const int N=1024;const int inf=1edges... 阅读全文
摘要：
#include #include #include #include #include #include using namespace std;int gcd(int u,int v){ if(u10*180) difms=10*360-difms; if(d... 阅读全文
摘要：
题目关键是建模，之后就是简单地增广路。#include#include#include#include#includeusing namespace std;const int N=1024;const int inf=1q;void ford(){ int a[N],vis[... 阅读全文