HDOJ 1532 Drainage Ditches

最新推荐文章于 2025-09-06 09:33:29 发布

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最新推荐文章于 2025-09-06 09:33:29 发布

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原文链接：http://www.cnblogs.com/CKboss/p/3350882.html

本文详细介绍了Farmer John 如何通过构建复杂的排水沟网络系统，有效解决暴雨导致的农田积水问题。文章深入探讨了排水沟的设计布局、水流控制以及如何最大效率地将水排出农田，确保农作物不受水患影响。通过分析Farmer John 的创新解决方案，展示了农业灌溉与水利工程的巧妙结合，为农田管理提供了实用的参考。

同样的题目我竟然连续写了3遍 =。=lll 人艰不拆

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6974 Accepted Submission(s): 3295

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

Sample Output

50

Source

USACO 93

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

const int MaxV=10000,MaxE=10000;
const int INF=0x3f3f3f3f;

struct Edge
{
    int to,next,flow;
}E[MaxE];

int Adj[MaxV],dist[MaxV],Size,src,sink;  bool vis[MaxV];

void Init()
{
    Size=0;
    memset(Adj,-1,sizeof(Adj));
}

void Add_Edge(int u,int v,int c)
{
    E[Size].to=v; E[Size].next=Adj; E[Size].flow=c; Adj=Size++;
    E[Size].to=u; E[Size].next=Adj[v]; E[Size].flow=0; Adj[v]=Size++;
}

void bfs()
{
    memset(vis,false,sizeof(vis));
    memset(dist,0,sizeof(dist));
    queue<int> q;
    q.push(src); vis[src]=true;
    while(!q.empty())
    {
        int u=q.front(); q.pop();
        for(int i=Adj;~i;i=E.next)
        {
            int v=E.to;
            if(E.flow&&!vis[v])
            {
                dist[v]=dist+1;
                q.push(v);
                vis[v]=true;
            }
        }
    }
}

int dfs(int u,int delta)
{
    if(u==sink)
    {
        return delta;
    }
    else
    {
        int ret=0;
        for(int i=Adj;~i&δi=E.next)
        {
            int v=E.to;
            if(E.flow&&dist[v]==dist+1)
            {
                int dd=dfs(v,min(delta,E.flow));
                E.flow-=dd; E[i^1].flow+=dd;
                ret+=dd; delta-=dd;
            }
        }
        return ret;
    }
}

int maxflow()
{
    int ret=0;
    while(true)
    {
        bfs();
        if(!vis[sink]) return ret;
        ret+=dfs(src,INF);
    }
}

int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        Init();
        for(int i=0;i<n;i++)
        {
            int a,b,c;
            cin>>a>>b>>c;
            Add_Edge(a,b,c);
        }
        src=1;sink=m;
        printf("%d\n",maxflow());
    }
    return 0;
}