POJ1273 Drainage Ditches【最大流】【SAP】

题目链接：

http://poj.org/problem?id=1273

题目大意：

农民John的田里有M个池塘和N条水沟用来排水，池塘编号为1~M，1号池塘是所有水沟的源点，

M号池塘是水沟的汇点。给你N条水沟所连接的池塘和所能流过的水量，求整个水沟从源点到汇点

最多能流多少水。

思路：

很明显的求网络流最大流问题。用链式前向星(邻接表)来存储网络，这样就不用考虑重边问题了。这

里的重边其实就是平行边。用SAP算法+GAP优化来求最大流就可以了。SAP+GAP模板参考我的另

一篇博文：http://blog.csdn.net/lianai911/article/details/44962653

AC代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 220;
const int MAXM = MAXN*MAXN;
const int INF = 0xffffff0;

struct EdgeNode
{
    int to;
    int w;
    int next;
}Edges[MAXM];
int Head[MAXN],id;

void AddEdges(int u,int v,int w)
{
    Edges[id].to = v;
    Edges[id].w = w;
    Edges[id].next = Head[u];
    Head[u] = id++;
    Edges[id].to = u;
    Edges[id].w = 0;
    Edges[id].next = Head[v];
    Head[v] = id++;
}

int Numh[MAXN],h[MAXN],curedges[MAXN],pre[MAXN];

void BFS(int end,int N)
{
    memset(Numh,0,sizeof(Numh));
    for(int i = 1; i <= N; ++i)
        Numh[h[i]=N]++;
    h[end] = 0;
    Numh[N]--;
    Numh[0]++;
    queue<int> Q;
    Q.push(end);

    while(!Q.empty())
    {
        int v = Q.front();
        Q.pop();
        int i = Head[v];
        while(i != -1)
        {
            int u = Edges[i].to;

            if(h[u] < N)
            {
                i = Edges[i].next;
                continue;
            }

            h[u] = h[v] + 1;
            Numh[N]--;
            Numh[h[u]]++;
            Q.push(u);
            i = Edges[i].next;
        }
    }
}

int SAP(int start,int end,int N)
{
    int Curflow,FlowAns = 0,temp,neck;
    memset(h,0,sizeof(h));
    memset(pre,-1,sizeof(pre));
    for(int i = 1; i <= N; ++i)
        curedges[i] = Head[i];

    BFS(end,N);
    int u = start;
    while(h[start] < N)
    {
        if(u == end)
        {
            Curflow = INF;
            for(int i = start; i != end; i = Edges[curedges[i]].to)
            {
                if(Curflow > Edges[curedges[i]].w)
                {
                    neck = i;
                    Curflow = Edges[curedges[i]].w;
                }
            }
            for(int i = start; i != end; i = Edges[curedges[i]].to)
            {
                temp = curedges[i];
                Edges[temp].w -= Curflow;
                Edges[temp^1].w += Curflow;
            }
            FlowAns += Curflow;
            u = neck;
        }
        int i;
        for(i = curedges[u]; i != -1; i = Edges[i].next)
            if(Edges[i].w && h[u]==h[Edges[i].to]+1)
                break;
        if(i != -1)
        {
            curedges[u] = i;
            pre[Edges[i].to] = u;
            u = Edges[i].to;
        }
        else
        {
            if(0 == --Numh[h[u]])
                break;
            curedges[u] = Head[u];
            for(temp = N,i = Head[u]; i != -1; i = Edges[i].next)
                if(Edges[i].w)
                    temp = min(temp,h[Edges[i].to]);
            h[u] = temp + 1;
            ++Numh[h[u]];
            if(u != start)
                u = pre[u];
        }
    }
    return FlowAns;
}

int main()
{
    int N,M,u,v,w;
    while(~scanf("%d%d",&N,&M))
    {
        memset(Head,-1,sizeof(Head));
        id = 0;
        for(int i = 0; i < N; ++i)
        {
            scanf("%d%d%d",&u,&v,&w);
            AddEdges(u,v,w);
        }
        printf("%d\n",SAP(1,M,M));
    }

    return 0;
}