HDU4280：Island Transport（最大流）

Island Transport

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 13187    Accepted Submission(s): 4156

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4280

Description:

　　In the vast waters far far away, there are many islands. People are living on the islands, and all the transport among the islands relies on the ships.
　　You have a transportation company there. Some routes are opened for passengers. Each route is a straight line connecting two different islands, and it is bidirectional. Within an hour, a route can transport a certain number of passengers in one direction. For safety, no two routes are cross or overlap and no routes will pass an island except the departing island and the arriving island. Each island can be treated as a point on the XY plane coordinate system. X coordinate increase from west to east, and Y coordinate increase from south to north.
　　The transport capacity is important to you. Suppose many passengers depart from the westernmost island and would like to arrive at the easternmost island, the maximum number of passengers arrive at the latter within every hour is the transport capacity. Please calculate it.

Input:

　　The first line contains one integer T (1<=T<=20), the number of test cases.
　　Then T test cases follow. The first line of each test case contains two integers N and M (2<=N,M<=100000), the number of islands and the number of routes. Islands are number from 1 to N.
　　Then N lines follow. Each line contain two integers, the X and Y coordinate of an island. The K-th line in the N lines describes the island K. The absolute values of all the coordinates are no more than 100000.
　　Then M lines follow. Each line contains three integers I1, I2 (1<=I1,I2<=N) and C (1<=C<=10000) . It means there is a route connecting island I1 and island I2, and it can transport C passengers in one direction within an hour.
　　It is guaranteed that the routes obey the rules described above. There is only one island is westernmost and only one island is easternmost. No two islands would have the same coordinates. Each island can go to any other island by the routes.

Output:

　　For each test case, output an integer in one line, the transport capacity.

Sample Input:

2 5 7 3 3 3 0 3 1 0 0 4 5 1 3 3 2 3 4 2 4 3 1 5 6 4 5 3 1 4 4 3 4 2 6 7 -1 -1 0 1 0 2 1 0 1 1 2 3 1 2 1 2 3 6 4 5 5 5 6 3 1 4 6 2 5 5 3 6 4

Sample Output:

9 6

题意：
给出n个点的坐标，然后会给出点与点之间的连接关系以及边的权值，问从最左边的点到最右边的点经过边的权值和最大为多少。

题解：
这道题和普通最大流的区别就是，这里是双向边，所以只需要稍微变一下就可以了。

在以前单向边的图中，我们建立了容量为0的反向边，之后会增加反向边的容量以便于进行”反悔操作“。

这里也同样，但是建反向边时容量与正向边相等，可以想一想为什么。只要理解了最大流的算法应该就比较好想。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#define INF 99999999
using namespace std;
typedef long long ll;
const int N = 1e5+ ;
int T,s,t,u,v,tot,n,m;
int head[N],d[N];
struct Edge{
    int v,next,c;
}e[N<<];
void adde(int u,int v,int c){
    e[tot].v=v;e[tot].next=head[u];e[tot].c=c;head[u]=tot++;
    e[tot].v=u;e[tot].next=head[v];e[tot].c=c;head[v]=tot++;
}
int bfs(){
    memset(d,,sizeof(d));d[s]=;
    queue<int> q;q.push(s);
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=head[u];i!=-;i=e[i].next){
            int v=e[i].v;
            if(!d[v] &&e[i].c>){
                d[v]=d[u]+;
                q.push(v);
            }
        }
    }
    return d[t]!=;
}
int dfs(int u,int a){
    if(u==t || a==) return a;
    int flow=,f;
    for(int i=head[u];i!=-;i=e[i].next){
        int v=e[i].v;
        if(d[v]!=d[u]+) continue ;
        f=dfs(v,min(a,e[i].c));
        if(f>){
            flow+=f;
            e[i].c-=f;
            e[i^].c+=f;
            a-=f;
            if(a==) break ;
        }
    }
    if(!flow) d[u]=-;
    return flow;
}
int Dinic(){
    int flow= ;
    while(bfs()) flow+=dfs(s,INF);
    return flow;
}
int main(){
    scanf("%d",&T);
    while(T--){
        tot=;memset(head,-,sizeof(head));
        scanf("%d%d",&n,&m);
        int minx = ,maxx = - ;
        for(int i=,u,v;i<=n;i++){
            scanf("%d%d",&u,&v);
            if(minx>u){
                minx=u;s=i;
            }
            if(maxx<u){
                maxx=u;t=i;
            }
        }
        for(int i=,u,v,c;i<=m;i++){
            scanf("%d%d%d",&u,&v,&c);
            adde(u,v,c);
        }
        printf("%d\n",Dinic());
    }
    return ;
}