1273 Drainage Ditches 网络流 SAP

最新推荐文章于 2021-07-19 20:05:53 发布

原创最新推荐文章于 2021-07-19 20:05:53 发布 · 430 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#sap #网络 #each #output #path #integer

acm_网络流专栏收录该内容

67 篇文章

订阅专栏

本文介绍了一种解决农场中排水沟渠流量最大化问题的方法。通过建立一套复杂的沟渠网络，并利用水流调控装置来确保水能从积水处快速排出，避免农作物受损。文中详细解释了如何通过算法计算最大排水速率。

Drainage Ditches

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21603		Accepted: 7723

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

Sample Output

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int inf=(1<<31)-1;
const int point_num=300;
int cap[point_num][point_num],dist[point_num],gap[point_num];//初始化见main里面 
int s0,t0,n;//源,汇和点数 
int find_path(int p,int limit=0x3f3f3f3f)
{
    if(p==t0)   return limit;
    for(int i=0;i<n;i++)
    if(dist[p]==dist[i]+1  &&  cap[p][i]>0)
    {
       int t=find_path(i,min(cap[p][i],limit));
       if(t<0)   return t;
       if(t>0)
       {
            cap[p][i]-=t;
            cap[i][p]+=t;
            return t;
       }
    }
    int label=n;
    for(int i=0;i<n;i++)  if(cap[p][i]>0)  label=min(label,dist[i]+1);
    if(--gap[dist[p]]==0  ||  dist[s0]>=n )   return -1;
    ++gap[dist[p]=label];
    return 0;
}
int sap()
{
    //初始化s,t
    s0=0,t0=n-1; 
    int t=0,maxflow=0;
    gap[0]=n;
    while((t=find_path(s0))>=0) maxflow+=t;
    return maxflow;
}
int main()
{
    int ci;
    while(cin>>ci>>n)
    {
        //初始化  
        memset(cap,0,sizeof(cap));
        memset(dist,0,sizeof(dist));
        memset(gap,0,sizeof(gap));
        //初始化cap
        while(ci--)//从0开始
        {
            int x,y,c;
            cin>>x>>y>>c;
            x--;y--;
            cap[x][y]+=c;//注意+= 
        }    
        int ans=sap();
        cout<<ans<<endl;
    }    
    return 0;
}