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(网络流 模板 Dinic) Drainage Ditches --POJ --1273

时间:2015-08-12 13:04:23      阅读:167      评论:0      收藏:0      [点我收藏+]

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链接:

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

 

代码:

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;

const int MAXN = 1005;
const int oo = 1e9+7;

struct Edge
{
    int v, flow, next;
}edge[MAXN];
int Head[MAXN], cnt;
int Layer[MAXN];

void InIt()
{
    cnt = 0;
    memset(Head, -1, sizeof(Head));
}


void AddEdge(int u, int v, int flow)
{
    edge[cnt].v = v;
    edge[cnt].flow = flow;
    edge[cnt].next = Head[u];
    Head[u] = cnt++;

    swap(u, v);

    edge[cnt].v = v;
    edge[cnt].flow = 0;
    edge[cnt].next = Head[u];
    Head[u] = cnt++;

}


bool BFS(int start, int End)   // 分层
{
    memset(Layer, 0, sizeof(Layer));
    queue<int> Q;
    Q.push(start);
    Layer[start] = 1;

    while(Q.size())
    {
        int u = Q.front();Q.pop();

        if(u == End)return true;

        for(int j=Head[u]; j!=-1; j=edge[j].next)
        {
            int v = edge[j].v;

            if(Layer[v] == 0 && edge[j].flow) // 判断Layer[v]==0,可以减少循环
            {
                Layer[v] = Layer[u] + 1;
                Q.push(v);
            }
        }
    }

    return false;
}


int DFS(int u, int MaxFlow, int End)  // DFS返回的是这条路上的最大流(最小值)
{
    if(u == End)return MaxFlow;

    int uflow = 0;

    for(int j=Head[u]; j!=-1; j=edge[j].next)
    {
        int v = edge[j].v;

        if(Layer[v]==Layer[u]+1 && edge[j].flow)
        {
            int flow = min(edge[j].flow, MaxFlow-uflow);
            flow = DFS(v, flow, End);

            edge[j].flow -= flow;
            edge[j^1].flow += flow;
            uflow += flow;

            if(uflow == MaxFlow)
                break;
        }
    }

    if(uflow == 0)
        Layer[u] = 0;

    return uflow;
}

int Dinic(int start, int End)
{
    int MaxFlow = 0;

    while(BFS(start, End) == true)  // 先找出一条从Start到End的路
        MaxFlow += DFS(start, oo, End); 

    return MaxFlow;
}

int main()
{
    int N, M;

    while(scanf("%d%d", &M, &N) != EOF)
    {
        int u, v, flow;

        InIt();

        while(M--)
        {
            scanf("%d%d%d", &u, &v, &flow);
            AddEdge(u, v, flow);
        }

        printf("%d\n", Dinic(1, N));
    }

    return 0;
}

 

(网络流 模板 Dinic) Drainage Ditches --POJ --1273

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原文地址:http://www.cnblogs.com/YY56/p/4723902.html

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