Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14375    Accepted Submission(s): 6810

Problem Description

Every
time it rains on Farmer John‘s fields, a pond forms over Bessie‘s
favorite clover patch. This means that the clover is covered by water
for awhile and takes quite a long time to regrow. Thus, Farmer John has
built a set of drainage ditches so that Bessie‘s clover patch is never
covered in water. Instead, the water is drained to a nearby stream.
Being an ace engineer, Farmer John has also installed regulators at the
beginning of each ditch, so he can control at what rate water flows into
that ditch.
Farmer John knows not only how many gallons of water
each ditch can transport per minute but also the exact layout of the
ditches, which feed out of the pond and into each other and stream in a
potentially complex network.
Given all this information, determine
the maximum rate at which water can be transported out of the pond and
into the stream. For any given ditch, water flows in only one direction,
but there might be a way that water can flow in a circle.

Input

The
input includes several cases. For each case, the first line contains
two space-separated integers, N (0 <= N <= 200) and M (2 <= M
<= 200). N is the number of ditches that Farmer John has dug. M is
the number of intersections points for those ditches. Intersection 1 is
the pond. Intersection point M is the stream. Each of the following N
lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei
<= M) designate the intersections between which this ditch flows.
Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <=
10,000,000) is the maximum rate at which water will flow through the
ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

Sample Output

50

网络流入门题，理解剩余网络，反向边,求源点到汇点最大流,网络流推荐博客：http://www.cnblogs.com/zsboy/archive/2013/01/27/2878810.html

不过这个算法(Edmond-Karp)的速度不算太快，最快的是DINIC算法,见下面。

#include <iostream>
#include <queue>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
const int N = 205;
const int INF = 999999999;
int flow[N]; ///标记从源点到当前节点实际还剩多少流量可用
int cap[N][N]; ///记录残留网络的容量
int pre[N]; ///记录前结点，为寻找路径做准备
int n,m;

int BFS(int src,int des)
{
    queue<int> q;
    for(int i=1; i<=m; i++)
    {
        pre[i]=-1;
    }
    pre[src] = 0;
    flow[src] = INF;
    q.push(src);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        if(u==des) break;
        for(int i=1; i<=m; i++)
        {
            if(i!=src&&cap[u][i]>0&&pre[i]==-1)
            {
                pre[i] = u;
                flow[i] = min(cap[u][i],flow[u]);
                q.push(i);
            }
        }
    }
    if(pre[des]==-1) return -1;///没有增广路了
    return flow[des];
}
void max_flow(int src,int des)
{
    int increaseRoad = 0;
    int MAXFLOW = 0;
    while((increaseRoad=BFS(src,des))!=-1)
    {
        int k = des;
        while(k!=src)
        {
            cap[pre[k]][k]-=increaseRoad; ///正向边的容量
            cap[k][pre[k]]+=increaseRoad; ///反向边的容量
            k = pre[k];
        }
        MAXFLOW+=increaseRoad;
    }
    printf("%d\n",MAXFLOW);
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(cap,0,sizeof(cap));
        for(int i=0; i<n; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            if(u==v) continue;
            cap[u][v]+=w;
        }
        max_flow(1,m);
    }
    return 0;
}

Dinic算法模板:http://blog.csdn.net/wall_f/article/details/8207595

#include <iostream>
#include <queue>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
const int N = 205;
const int INF = 999999999;
struct Edge{
    int v,w,next;
}edge[N*N];
int head[N];
int level[N];
int n,m;
void addEdge(int u,int v,int w,int &k){
    edge[k].v = v,edge[k].w=w,edge[k].next=head[u],head[u]=k++;
    edge[k].v = u,edge[k].w=0,edge[k].next=head[v],head[v]=k++; ///建立反向边
}
int BFS(int src,int des){
    queue<int >q;
    memset(level,0,sizeof(level));
    level[src]=1;
    q.push(src);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        if(u==des) return 1;
        for(int k = head[u];k!=-1;k=edge[k].next){
            int v = edge[k].v,w=edge[k].w;
            if(level[v]==0&&w!=0){
                level[v]=level[u]+1;
                q.push(v);
            }
        }
    }
    return -1;
}
int dfs(int u,int des,int increaseRoad){
    if(u==des) return increaseRoad;
    int ret=0;
    for(int k=head[u];k!=-1;k=edge[k].next){
        int v = edge[k].v,w=edge[k].w;
        if(level[v]==level[u]+1&&w!=0){
            int MIN = min(increaseRoad-ret,w);
            w = dfs(v,des,MIN);
            edge[k].w -=w;
            edge[k^1].w+=w;
            ret+=w;
            if(ret==increaseRoad) return ret;
        }
    }
    return ret;
}
int Dinic(int src,int des){
    int ans = 0;
    while(BFS(src,des)!=-1) ans+=dfs(src,des,INF);
    return ans;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(head,-1,sizeof(head));
        int tot = 0;
        for(int i=0;i<n;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            if(u==v) continue;
            addEdge(u,v,w,tot);
        }
        printf("%d\n",Dinic(1,m));
    }
    return 0;
}