最小割

题目描述
给定\(n\)个点\(m\)条边的有向图，删掉第\(i\)条边的代价为\(c[i]\)。

请删掉代价之和最少的边，使得从\(1\)号点出发到达不了\(n\)号点。
输入
第一行包含一个正整数\(T(1\leq T\leq 10)\)，表示测试数据的组数。

每组数据第一行包含两个正整数\(n,m(1\leq n\leq 50,1\leq m\leq 1000)\)。

接下来\(m\)行，每行三个正整数\(a[i],b[i],c[i](1\leq a[i],b[i]\leq n,a[i]\neq b[i],1\leq c[i]\leq 10000)\)，表示一条起点是\(a[i]\)，终点是\(b[i]\)的边，删掉它的代价是\(c[i]\)。
输出
对于每组数据输出一行一个整数，即最小代价。

#include<bits/stdc++.h>
using namespace std;
typedef vector<string> VS;
const int N = 110,INF = 1e9;
struct Edge {
  int from, to, cap, flow;
  Edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};

bool operator < (const Edge& a, const Edge& b) {
  return a.from < b.from || (a.from == b.from && a.to < b.to);
}

struct ISAP {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[N];
  bool vis[N];
  int d[N],cur[N],p[N],num[N];

  void AddEdge(int from, int to, int cap) {
    edges.push_back(Edge(from, to, cap, 0));
    edges.push_back(Edge(to, from, 0, 0));
    m = edges.size();
    G[from].push_back(m - 2);
    G[to].push_back(m - 1);
  }

  bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(t);
    vis[t] = 1;
    d[t] = 0;
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i] ^ 1];
        if (!vis[e.from] && e.cap > e.flow) {
          vis[e.from] = 1;
          d[e.from] = d[x] + 1;
          Q.push(e.from);
        }
      }
    }
    return vis[s];
  }

  void init(int n) {
    this->n = n;
    for (int i = 0; i < n; i++) G[i].clear();
    edges.clear();
  }

  int Augment() {
    int x = t, a = INF;
    while (x != s) {
      Edge& e = edges[p[x]];
      a = min(a, e.cap - e.flow);
      x = edges[p[x]].from;
    }
    x = t;
    while (x != s) {
      edges[p[x]].flow += a;
      edges[p[x] ^ 1].flow -= a;
      x = edges[p[x]].from;
    }
    return a;
  }

  int Maxflow(int s, int t) {
    this->s = s;
    this->t = t;
    int flow = 0;
    BFS();
    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; i++) num[d[i]]++;
    int x = s;
    memset(cur, 0, sizeof(cur));
    while (d[s] < n) {
      if (x == t) {
        flow += Augment();
        x = s;
      }
      int ok = 0;
      for (int i = cur[x]; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if (e.cap > e.flow && d[x] == d[e.to] + 1) {
          ok = 1;
          p[e.to] = G[x][i];
          cur[x] = i;
          x = e.to;
          break;
        }
      }
      if (!ok) {
        int m = n - 1;
        for (int i = 0; i < G[x].size(); i++) {
          Edge& e = edges[G[x][i]];
          if (e.cap > e.flow) m = min(m, d[e.to]);
        }
        if (--num[d[x]] == 0) break;
        num[d[x] = m + 1]++;
        cur[x] = 0;
        if (x != s) x = edges[p[x]].from;
      }
    }
    return flow;
  }
};
ISAP FG;
void solve()
{
	int n,m;
	cin>>n>>m;
	//最小割的源点和汇点直接选用图里的点即可
	int s = 1,t = n;
	FG.init(n+1);
	
	for(int i=0;i<m;++i)
	{
		int u,v,w;
		cin>>u>>v>>w;
		FG.AddEdge(u,v,w);
	}
	cout<<FG.Maxflow(s,t)<<'\n';
}
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		solve();
	}
}