zoukankan      html  css  js  c++  java
  • HDOJ1532||POJ1273 Drainage Ditches[最大流模版]

    Drainage Ditches

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 4515    Accepted Submission(s): 2108


    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
     
    Source
     
    Recommend
    lwg
     
     
     
     
     
     
     
    模版,水过。
    code:
     1 #include <iostream>   
     2 #include <iomanip>   
     3 #include <fstream>   
     4 #include <sstream>   
     5 #include <algorithm>   
     6 #include <string>   
     7 #include <set>   
     8 #include <utility>   
     9 #include <queue>   
    10 #include <stack>   
    11 #include <list>   
    12 #include <vector>   
    13 #include <cstdio>   
    14 #include <cstdlib>   
    15 #include <cstring>   
    16 #include <cmath>   
    17 #include <ctime>   
    18 #include <ctype.h> 
    19 using namespace std;
    20 
    21 #define MAXN 500
    22 #define intmax 99999999
    23 
    24 int cap[MAXN][MAXN];
    25 int flow[MAXN][MAXN];
    26 int a[MAXN];
    27 int pre[MAXN];
    28 int n,m;
    29 
    30 
    31 int EK()
    32 {
    33     queue<int>Que;
    34     int f=0;
    35     memset(flow,0,sizeof(flow));
    36     for(;;)
    37     {
    38         memset(a,0,sizeof(a));
    39         a[1]=intmax;
    40         Que.push(1);
    41         while(!Que.empty())
    42         {
    43             int u=Que.front();
    44             Que.pop();
    45             for(int i=1;i<=m;i++)
    46                 if(!a[i]&&cap[u][i]>flow[u][i])
    47                 {
    48                     pre[i]=u;
    49                     Que.push(i);
    50                     a[i]=a[u]>cap[u][i]-flow[u][i]?cap[u][i]-flow[u][i]:a[u];
    51                 }
    52         }
    53         if(a[m]==0)
    54             break;
    55         for(int i=m;i!=1;i=pre[i])
    56         {
    57             flow[pre[i]][i]+=a[m];
    58             flow[i][pre[i]]-=a[m];
    59         }
    60         f+=a[m];
    61     }
    62     return f;
    63 }
    64 
    65 int main()
    66 {
    67     int a,b,temp;
    68     while(~scanf("%d%d",&n,&m))
    69     {
    70         memset(cap,0,sizeof(cap));
    71         for(int i=0;i<n;i++)
    72         {
    73             scanf("%d%d%d",&a,&b,&temp);
    74             cap[a][b]+=temp;
    75         }
    76         printf("%d\n",EK());
    77     }
    78     return 0;
    79 }
  • 相关阅读:
    xml学习_上篇
    回首博客 年满一周年,记录90个点滴瞬间
    编写PHP代码总结
    代码高亮插件SyntaxHighlighter
    记忆,是一座沙城!
    dreamweaver中的 map怎么调用?_制作热点图像区域
    手机端rem如何适配_rem详解及使用方法
    彻底弄懂css中单位px和em,rem的区别
    JavaScript学习方法
    10条建议让你创建更好的jQuery插件
  • 原文地址:https://www.cnblogs.com/XBWer/p/2639796.html
Copyright © 2011-2022 走看看