minute ostream 代碼 gin ted pat 當前 earch http

Drainage Ditches

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 85250		Accepted: 33164

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

USACO 93 題意：最大流問題 代碼：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
#define MAX 205
#define INF 0x3f3f3f3f
#define pb push_back
using namespace std;
struct edge{
int to,cap,rev;
edge(){}
edge(int to,int cap,int rev):to(to),cap(cap),rev(rev){}
};
vector<edge>G[MAX];
bool used[MAX];
void add_edge(int from,int to,int cap)
{
    G[from].pb(edge(to,cap,G[to].size()));
    G[to].pb(edge(from,0,G[from].size()-1));
}
int dfs(int v,int t,int f)
{
    if(v==t)return f;
    used[v]=true;
    for(int i=0;i<G[v].size();i++)
    {
        edge &e=G[v][i];
        //cout<<e.to<<endl;
        if(!used[e.to]&&e.cap>0)
        {
            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;
}
int max_flow(int s,int t)
{
    int flow=0;
    for(;;)
    {
        memset(used,0,sizeof(used));
        int f=dfs(s,t,INF);
        if(f==0)
            return flow;
        flow+=f;
    }
}
int main()
{
    int n,m;
       while(cin>>n>>m){
           for(int i=0;i<n;i++)
            G[i].clear();
        for(int i=0;i<n;i++)
        {
            int u,v,cap;
            cin>>u>>v>>cap;
            add_edge(u,v,cap);
        }
        cout<<max_flow(1,m)<<endl;
    }
}

Ford_Fulkerson

#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 205;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int from,to,cap,flow;
    Edge(){}
    Edge(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow){}
};

struct Dinic
{
    int n,m,s,t;            //結點數,邊數(包括反向弧),源點與匯點編號
    vector<Edge> edges;     //邊表 edges[e]和edges[e^1]互為反向弧
    vector<int> G[maxn];    //鄰接表,G[i][j]表示結點i的第j條邊在e數組中的序號
    bool vis[maxn];         //BFS使用,標記一個節點是否被遍歷過
    int d[maxn];            //d[i]表從起點s到i點的距離(層次)
    int cur[maxn];          //cur[i]表當前正訪問i節點的第cur[i]條弧

    void init(int n,int s,int t)
    {
        this->n=n,this->s=s,this->t=t;
        for(int i=0;i<=n;i++) G[i].clear();
        edges.clear();
    }

    void AddEdge(int from,int to,int cap)
    {
        edges.push_back( Edge(from,to,cap,0) );
        edges.push_back( Edge(to,from,0,0) );
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BFS()
    {
        memset(vis,0,sizeof(vis));
        queue<int> Q;//用來保存節點編號的
        Q.push(s);
        d[s]=0;
        vis[s]=true;
        while(!Q.empty())
        {
            int x=Q.front(); Q.pop();
            for(int i=0; i<G[x].size(); i++)
            {
                Edge& e=edges[G[x][i]];
                if(!vis[e.to] && e.cap>e.flow)
                {
                    vis[e.to]=true;
                    d[e.to] = d[x]+1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    //a表示從s到x目前為止所有弧的最小殘量
    //flow表示從x到t的最小殘量
    int DFS(int x,int a)
    {
        if(x==t || a==0)return a;
        int flow=0,f;//flow用來記錄從x到t的最小殘量
        for(int& i=cur[x]; i<G[x].size(); i++)
        {
            Edge& e=edges[G[x][i]];
            if(d[x]+1==d[e.to] && (f=DFS( e.to,min(a,e.cap-e.flow) ) )>0 )
            {
                e.flow +=f;
                edges[G[x][i]^1].flow -=f;
                flow += f;
                a -= f;
                if(a==0) break;
            }
        }
        if(!flow) d[x] = -1;///炸點優化
        return flow;
    }

    int Maxflow()
    {
        int flow=0;
        while(BFS())
        {
            memset(cur,0,sizeof(cur));
            flow += DFS(s,INF);
        }
        return flow;
    }
}Di;
int main()
{
    int n,m;

    while(cin>>n>>m){
        Di.init(n,1,m);
    for(int i=0;i<n;i++)
    {
        int v,u,rap;
        cin>>u>>v>>rap;
        Di.AddEdge(u,v,rap);
    }
     cout<<Di.Maxflow()<<endl;
   }
}

Dinic

poj Drainage Ditches（最大流入門）