zoukankan      html  css  js  c++  java
  • POJ1273 Drainage Ditches (网络流)

                                                             Drainage Ditches
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 69983   Accepted: 27151

    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
    【分析】我直接套的树上的网络流标号法模板。
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <time.h>
    #include <string>
    #include <map>
    #include <stack>
    #include <vector>
    #include <set>
    #include <queue>
    #define inf 0x3f3f3f3f
    #define mod 1000000007
    typedef long long ll;
    using namespace std;
    const int N=205;
    const int M=1005;
    int s,t,n,m,num,k;
    int customer[N][N];
    int flow[N][N];
    int house[M],last[M];
    int pre[N],minflow[N];
    
    void BFS() {
        queue<int>q;
        int p=0;
        memset(flow,0,sizeof(flow));
        minflow[1]=inf;
        while(1) {
            while(!q.empty())q.pop();
            for(int i=0; i<N; i++)pre[i]=-2;
            pre[1]=-1;
            q.push(1);
            while(!q.empty()&&pre[t]==-2) {
                int v=q.front();
                q.pop();
                for(int i=1; i<t+1; i++) {
                    if(pre[i]==-2&&(p=customer[v][i]-flow[v][i])) {
                        pre[i]=v;
                        q.push(i);
                        minflow[i]=min(p,minflow[v]);
                    }
                }
            }
            if(pre[t]==-2)break;
            int j;
            for(int i=pre[t],j=t; i>=0; j=i,i=pre[i]) {
                flow[i][j]+=minflow[t];
                flow[j][i]=-flow[i][j];
            }
        }
        for(int i=0; i<t; i++)p+=flow[i][t];
        printf("%d
    ",p);
    }
    int main() {
        while(~scanf("%d%d",&m,&n)) {
            memset(last,0,sizeof(last));
            memset(customer,0,sizeof(customer));
            s=1;
            t=n;
            int a,b,c;
            for(int i=1; i<=m; i++){
                scanf("%d%d%d",&a,&b,&c);
                customer[a][b]+=c;
            }
            BFS();
        }
        return 0;
    }
    View Code
  • 相关阅读:
    【Codeforces 923A】Primal Sport
    【Codeforces 924C】Riverside Curio
    【Codeforces 682C】Alyona and the Tree
    【Codeforces 1118D1】Coffee and Coursework (Easy version)
    【Codeforces 493C】Vasya and Basketball
    【Codeforces 598D】Igor In the Museum
    js 时间格式化
    C#自定义规则对比两个集合的对象是否相等
    VUE的组件DEMO
    js 去掉指定符号的字符串做法
  • 原文地址:https://www.cnblogs.com/jianrenfang/p/5826299.html
Copyright © 2011-2022 走看看