hdu5889 Barricade（最短路+网络流）

思路：先跑一次最短路，然后把最短路上的边扔进网络流跑一次最小割就好了，和HDU3416异曲同工

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2005;
#define inf 1e9
#define INF 1e9
const int maxm = 4000000+50;
struct Node
{
	int v,w;
	Node(int vv,int ww):v(vv),w(ww){};
};
vector<Node>e[maxn];
int s,t,n,m,vs,vt;
int d[maxn];
int vis[maxn];
void spfa()
{
	memset(vis,0,sizeof(vis));
	for(int i = 1;i<=n;i++)
		d[i]=inf;
	d[s]=0;
	queue<int>q;
	q.push(s);
    while(!q.empty())
	{
        int u = q.front();
		q.pop();
		vis[u]=0;
		for(int i = 0;i<e[u].size();i++)
		{
			int v = e[u][i].v;
			int w = e[u][i].w;
			if(d[v]>d[u]+1)
			{
				d[v]=d[u]+1;
				if(!vis[v])
					q.push(v);
				vis[v]=1;
			}
		}
	}
}


struct Edge
{
	int from,to,cap,flow;
	Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct Dinic
{
//	int n,m;
    int s,t;
	vector<Edge>edges;        //边数的两倍
	vector<int> G[maxn];      //邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
	bool vis[maxn];           //BFS使用
	int d[maxn];              //从起点到i的距离
	int cur[maxn];            //当前弧下标
	void init()
	{
	   for (int i=0;i<=n+1;i++)
		   G[i].clear();
	   edges.clear();
	}
	void AddEdge(int from,int to,int cap)
	{
		edges.push_back(Edge(from,to,cap,0));
		edges.push_back(Edge(to,from,0,0));        //反向弧
		int mm=edges.size();
		G[from].push_back(mm-2);
		G[to].push_back(mm-1);
	}
	bool BFS()
	{
		memset(vis,0,sizeof(vis));
		queue<int>q;
		q.push(s);
		d[s]=0;
		vis[s]=1;
		while (!q.empty())
		{
			int x = q.front();q.pop();
			for (int i = 0;i<G[x].size();i++)
			{
				Edge &e = edges[G[x][i]];
				if (!vis[e.to] && e.cap > e.flow)
				{
					vis[e.to]=1;
					d[e.to] = d[x]+1;
					q.push(e.to);
				}
			}
		}
		return vis[t];
	}

	int DFS(int x,int a)
	{
		if (x==t || a==0)
			return a;
		int flow = 0,f;
		for(int &i=cur[x];i<G[x].size();i++)
		{
			Edge &e = edges[G[x][i]];
			if (d[x]+1 == d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow)))>0)
			{
				e.flow+=f;
				edges[G[x][i]^1].flow-=f;
				flow+=f;
				a-=f;
				if (a==0)
					break;
			}
		}
		return flow;
	}

	int Maxflow(int s,int t)
	{
		this->s=s;
		this->t=t;
		int flow = 0;
		while (BFS())
		{
			memset(cur,0,sizeof(cur));
			flow+=DFS(s,INF);
		}
		return flow;
	}
}dc;

int main()
{
    int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
        for(int i = 0;i<=n;i++)
			e[i].clear();
		for(int i = 1;i<=m;i++)
		{
			int u,v,di;
			scanf("%d%d%d",&u,&v,&di);
			e[u].push_back(Node(v,di));
			e[v].push_back(Node(u,di));
		}
		s=1,t=n;
		spfa();
		dc.init();
		for(int i = 1;i<=n;i++)
			for(int j = 0;j<e[i].size();j++)
				if(d[e[i][j].v]==d[i]+1)
					dc.AddEdge(i,e[i][j].v,e[i][j].w);
		printf("%d\n",dc.Maxflow(s,t));
	}
}

Problem Description

The empire is under attack again. The general of empire is planning to defend his castle. The land can be seen as N towns and M roads, and each road has the same length and connects two towns. The town numbered 1 is where general's castle is located, and the town numbered N is where the enemies are staying. The general supposes that the enemies would choose a shortest path. He knows his army is not ready to fight and he needs more time. Consequently he decides to put some barricades on some roads to slow down his enemies. Now, he asks you to find a way to set these barricades to make sure the enemies would meet at least one of them. Moreover, the barricade on the i-th road requires wi units of wood. Because of lacking resources, you need to use as less wood as possible.

 

Input

The first line of input contains an integer t, then t test cases follow.
For each test case, in the first line there are two integers N(N≤1000) and M(M≤10000).
The i-the line of the next M lines describes the i-th edge with three integers u,v and w where 0≤w≤1000 denoting an edge between u and v of barricade cost w.

 

Output

For each test cases, output the minimum wood cost.

 

Sample Input

1
4 4
1 2 1
2 4 2
3 1 3
4 3 4

 

Sample Output

4