HDU 2435 There is a war （网络流-最小割）

There is a war

Problem Description

      There is a sea.

      There are N islands in the sea.

      There are some directional bridges connecting these islands.

      There is a country called Country One located in Island 1.

      There is another country called Country Another located in Island N. 

      There is a war against Country Another, which launched by Country One.

      There is a strategy which can help Country Another to defend this war by destroying the bridges for the purpose of making Island 1 and Island n disconnected.

      There are some different destroying costs of the bridges.

      There is a prophet in Country Another who is clever enough to find the minimum total destroying costs to achieve the strategy.

      There is an architecture in Country One who is capable enough to rebuild a bridge to make it unbeatable or build a new invincible directional bridge between any two countries from the subset of island 2 to island n-1.

      There is not enough time for Country One, so it can only build one new bridge, or rebuild one existing bridge before the Country Another starts destroying, or do nothing if happy.

      There is a problem: Country One wants to maximize the minimum total destroying costs Country Another needed to achieve the strategy by making the best choice. Then what’s the maximum possible result?

 

Input

      There are multiple cases in this problem.

      There is a line with an integer telling you the number of cases at the beginning.

      The are two numbers in the first line of every case, N(4<=N<=100) and M(0<=M<=n*(n-1)/2), indicating the number of islands and the number of bridges.

      There are M lines following, each one of which contains three integers a, b and c, with 1<=a, b<=N and 1<=c<=10000, meaning that there is a directional bridge from a to b with c being the destroying cost.

      There are no two lines containing the same a and b.

 

Output

      There is one line with one integer for each test case, telling the maximun possible result.

 

Sample Input

4
4 0
4 2
1 2 2
3 4 2
4 3
1 2 1
2 3 1
3 4 10
4 3
1 2 5
2 3 2
3 4 3

 

Sample Output

0
2
1
3

 

Source

2008 Asia Chengdu Regional Contest Online

 

题目大意：

n个岛通过有向边连在一起，countryOne坐落在1号岛上，countryAny坐落在n号岛上，如今1 要进攻 n ，n为了抵御1的攻击，要毁坏边，没条边毁坏要花费，如今1能够在 2-n 随意两个岛上建立一个摧毁不了的边，使得 countryAny 为了抵御进攻最小的花费最大为多少？

解题思路：

首先，最小割能够理解成网络流的流量，第一步 。countryAny肯定要用最小的花费使得图不连通，这个花费就是最小割。

可是，countryOne这个时候能够再建一条边使得图再次连通，所以得找出最小割的边集，边集就是  从 countryOne相邻的点 到 与countryOne不相邻的点 连接的边，

所以。再枚举这条边。加入到残余网络中。求全部答案中最大的与之前的最小割相加就是答案。

解题代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=100,maxm=10000;

struct edge{
    int u,v,f,next;
}e[maxm+10];

int src,sink,cnt,head[maxn+10];
int visited[maxn+10],marked;
int dist[maxn+10];
int n,m;

void adde(int u,int v,int f){
	e[cnt].u=u,e[cnt].v=v,e[cnt].f=f,e[cnt].next=head[u],head[u]=cnt++;
	e[cnt].u=v,e[cnt].v=u,e[cnt].f=0,e[cnt].next=head[v],head[v]=cnt++;
}

void bfs(){
    marked++;
    for(int i=0;i<=sink;i++) dist[i]=0;
    queue <int> q;
    visited[src]=marked;
    q.push(src);
    while(!q.empty()){
        int s=q.front();
        q.pop();
        for(int i=head[s];i!=-1;i=e[i].next){
            int d=e[i].v;
            if(e[i].f>0 && visited[d]!=marked ){
                q.push(d);
                dist[d]=dist[s]+1;
                visited[d]=marked;
            }
        }
    }
}

int dfs(int u,int delta){
    if(u==sink) return delta;
    else{
        int ret=0;
        for(int i=head[u];delta && i!=-1;i=e[i].next){
            if(e[i].f>0 && dist[e[i].v]==dist[u]+1){
                int d=dfs(e[i].v,min(e[i].f,delta));
                e[i].f-=d;
                e[i^1].f+=d;
                delta-=d;
                ret+=d;
            }
        }
        return ret;
    }
}

int maxflow(){
    int ret=0;
    while(true){
        bfs();
        if(visited[sink]!=marked) return ret;
        ret+=dfs(src,INF);
    }
    return ret;
}

void initial(){
    cnt=0;
    memset(head,-1,sizeof(head));
    marked++;
}

void input(){
    scanf("%d%d",&n,&m);
    src=1,sink=n;
    while(m-- >0){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        adde(u,v,w);
    }
}

void solve(){
    int ans=maxflow();
    bool f[maxn+10];
    memset(f,false,sizeof(f));
    queue <int> q;
    q.push(src);
    f[src]=true;
    while(!q.empty() ){
        int s=q.front();
        q.pop();
        for(int i=head[s];i!=-1;i=e[i].next){
            int d=e[i].v;
            if(e[i].f>0 && !f[d]){
                q.push(d);
                f[d]=true;
            }
        }
    }

    vector <edge> v;
    for(int i=0;i<cnt;i++) v.push_back(e[i]);

    int tmp=0;
    for(int i=2;i<=n-1;i++){
        for(int j=2;j<=n-1;j++){
            if( f[i]  &&  ( ! f[j] ) ){
                int headu=head[i],headv=head[j];

                adde(i,j,INF);
                int flow=maxflow();
                if(flow>tmp) tmp=flow;

                cnt-=2;
                head[i]=headu,head[j]=headv;
                for(int t=0;t<cnt;t++) e[t]=v[t];
            }
        }
    }
    cout<<ans+tmp<<endl;
}

int main(){
    int t;
    scanf("%d",&t);
    while(t-- >0){
        initial();
        input();
        solve();
    }
    return 0;
}