HDU5889 Barricade bfs+网络流裸模板 2016ICPC沈阳网络赛1011

沈阳预选赛的1011，刚开始没想到用bfs找最短路，还找了一个能找出全部多条最短路径的板子，结果一直犹豫最短路+网络流会不会超时，到了也没写完这题。

K - Barricade

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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  w_i  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 \leq 1000)  and  M(M \leq 10000) . 
The  i -the line of the next  M  lines describes the  i -th edge with three integers  u,v  and  w  where  0 \leq w \leq 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

题解：这里由于它全部路径的长度都为1，因此不须要什么算法去实现找多条最短路，只须要用bfs给点标号，由bfs的最优解特性可知，它找到的必定是最短路上的点，当搜完一遍后，用深搜再搜一遍，将全部知足dis[v]=dis[u]+1,的点建图造成一个新的网络，这样咱们就能够在新网络上裸最大流了，因为他要求用最少的木头去建造障碍物，实际上就是求这张网络带源点的点集和带汇点的点集之间的最小割，也就是最大流，这样就是裸模板了。

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <vector>
#include <queue>
#define MEMINF(a) memset(a,0x3f,sizeof a)
#define MEM(a,b) memset(a,b,sizeof a)
using namespace std;
typedef long long LL;
const int MAXN=1e5+10;
const int MAXM=4e5+10;
const int INF=0x3f3f3f3f;
struct MaxFlow{
    int head[MAXN];
    struct Edge{
        int u,v,nex,cap,flow;
    }edge[MAXM];
    int tot,s,t,n,m;
    int dis[MAXN];
    int cur[MAXN];
    void Dinic_init(int n,int m){
        this->s=n;
        this->t=1;
        this->n=n;
        this->m=m;
        MEM(head,-1);
        tot=0;
    }
    void Addedge(int u,int v,int w) {
        edge[tot].v=v,edge[tot].cap=w,edge[tot].flow=0,edge[tot].nex=head[u],head[u]=tot++;
        edge[tot].v=u,edge[tot].cap=0,edge[tot].flow=0,edge[tot].nex=head[v],head[v]=tot++;
    }
    bool bfs() {
        MEM(dis,-1);
        queue<int>q;
        dis[s]=0;
        q.push(s);
        while (!q.empty()) {
            int u=q.front();
            q.pop();
            for (int i=head[u]; ~i; i=edge[i].nex) {
                int v=edge[i].v;
                if (edge[i].cap>edge[i].flow&&dis[v]==-1) {
                    q.push(v);
                    dis[v]=dis[u]+1;
                }
            }
        }
        return dis[t]!=-1;
    }
    int dfs(int u,int delta) {
        if (u==t||delta==0)
            return delta;
            int ret=0;
            int aug;
            for(int &i=cur[u]; ~i; i=edge[i].nex) {
                int v=edge[i].v;
                if (dis[v]==dis[u]+1&&(aug=dfs(v,min(edge[i].cap-edge[i].flow,delta)))>0) {
                    edge[i].flow+=aug;
                    edge[i^1].flow-=aug;
                    delta-=aug;
                    ret+=aug;
                    if (delta==0) break;
                }
            }
            return ret;
    }
    void dinic() {
        int ret=0;
        while(bfs()) {
            memcpy(cur,head,sizeof head);
            ret+=dfs(s,INF);
            //cout<<ret<<endl;
        }
        cout<<ret<<endl;
     }
}nima;
struct Search_{
    int head[MAXN];
    int level[MAXN];
    int tot;
    int s;
    int vis[MAXN];
    struct Edge {
    int u,v,w;
    int nex;
    }edge[MAXM];
    void Addedge(int u,int v,int w) {
        edge[tot].v=v;edge[tot].u=u;edge[tot].w=w;edge[tot].nex=head[u],head[u]=tot++;
        edge[tot].v=u;edge[tot].u=v;edge[tot].w=w;edge[tot].nex=head[v],head[v]=tot++;
    }
    void search_init(int n) {
        MEM(level,-1);
        tot=0;
        MEM(head,-1);
        MEM(vis,0);
        this->s=n;
    }
    void bfs() {
        queue<int>q;
        level[s]=0;
        q.push(s);
        while (!q.empty()) {
            int u=q.front();
            q.pop();
            for (int i=head[u]; ~i; i=edge[i].nex) {
                int v=edge[i].v;
                if (level[v]==-1) {
                    level[v]=level[u]+1;
                    q.push(v);
            }
            }
        }
    }
    void dfs(int u) {
        vis[u]=1;
        for (int i=head[u]; ~i; i=edge[i].nex) {
            int v=edge[i].v;
            if (level[v]==level[u]+1) {
                //printf("u:%d v:%d w:%d\n",u,v,edge[i].w);
                nima.Addedge(u,v,edge[i].w);
                if (!vis[v])
                    dfs(v);
            }
        }
    }
    void search_work() {
        bfs();
        dfs(s);
        return;
    }
}cao;

int main(){
    int n,m;
    int Test;
    cin>>Test;
    while (Test--){
        scanf("%d %d",&n,&m);
        int u,v,w;
        cao.search_init(n);
        nima.Dinic_init(n,m);
        for (int i=0; i<m; ++i) {
            scanf("%d %d %d",&u,&v,&w);
            cao.Addedge(u,v,w);
        }
        cao.search_work();
        nima.dinic();

    }

}