HDU 1532 Drainage Ditches 最大网络流

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18483    Accepted Submission(s): 8752

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 41 2 401 4 202 4 202 3 303 4 10

 

Sample Output

50

 

Source

USACO 93

 

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lwg

 

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题意&思路：给出m条河流n个点及河流流量，求最大流，网络流基础题

AC代码：

#include<cstdio>#include<vector>#include<queue>#include<cstring>#define INF 1e9using namespace std;const int MAX_V=500;struct edge{    int to,cap,rev;     //终点，容量，反向边};vector<edge> G[MAX_V];  //图的邻接表表示int level[MAX_V];       //顶点到原点的距离标号int iter[MAX_V];        //当前弧，在其之前的边已经没有用了//增加一条从from到to的容量为cap的边void add_edge(int from,int to,int cap){    G[from].push_back((edge){to,cap,G[to].size()});    G[to].push_back((edge){from,0,G[from].size()-1});}//通过BFS计算从原点出发的距离标号void bfs(int s){    memset(level,-1,sizeof(level));    queue<int > que;    level[s]=0;    que.push(s);    while(!que.empty()){        int v=que.front(); que.pop();        for(int i=0;i<G[v].size();i++){            edge& e=G[v][i];            if(e.cap>0&&level[e.to]<0){                level[e.to]=level[v]+1;                que.push(e.to);            }        }    }}//通过DFS寻找增广路int dfs(int v,int t,int f){    if(v==t) return f;    for(int& i=iter[v]; i<G[v].size(); i++){        edge& e=G[v][i];        if(e.cap>0 && level[v]<level[e.to]){            int d=dfs(e.to,t,min(f,e.cap));            if(d>0){                e.cap -= d;                G[e.to][e.rev].cap+=d;                return d;            }        }    }    return 0;}//求解从s到t的最大流int max_flow(int s,int t){    int flow=0;    for(;;){        bfs(s);        if(level[t]<0) return flow;        memset(iter,0,sizeof(iter));        int f;        while((f=dfs(s,t,INF))>0){            flow+=f;        }    }}int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF){        for(int i=0;i<n;i++) G[i].clear(); //使用前记得清空        for(int i=0;i<n;i++){            int from,to,cap;            scanf("%d%d%d",&from,&to,&cap);            add_edge(from,to,cap);        }        int res=max_flow(1,m);        printf("%d\n",res);    }    return 0;}