POJ 1273 (基础最大流) Drainage Ditches

虽然算法还没有理解透，但以及迫不及待地想要A道题了。

非常裸的最大流，试试lrj的模板练练手。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

const int maxn = 200 + 10;
const int INF = 1000000000;

struct Edge
{
    int from, to, cap, flow;
    Edge(int u, int v, int c, int f):from(u), to(v), cap(c), flow(f) {}
};

int n, m, M;
vector<Edge> edges;
vector<int> G[maxn];
int a[maxn];    //到起点i的可改进量
int p[maxn];    //最短路数上p的入弧编号

void Init(int n)
{
    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));    //反向弧
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
}

int MaxFlow(int s, int t)
{
    int flow = 0;
    for(;;)
    {
        memset(a, 0, sizeof(a));
        queue<int> Q;
        Q.push(s);
        a[s] = INF;
        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(!a[e.to] && e.cap > e.flow)
                {
                    p[e.to] = G[x][i];
                    a[e.to] = min(a[x], e.cap-e.flow);
                    Q.push(e.to);
                }
            }
            if(a[t]) break;
        }
        if(!a[t]) break;
        for(int u = t; u != s; u = edges[p[u]].from)
        {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
        }
        flow += a[t];
    }
    return flow;
}

int main()
{
    freopen("in.txt", "r", stdin);

    while(scanf("%d%d", &M, &n) == 2)
    {
        Init(n);
        int u, v, c;
        for(int i = 0; i < M; ++i)
        {
            scanf("%d%d%d", &u, &v, &c);
            AddEdge(u-1, v-1, c);
        }
        printf("%d
", MaxFlow(0, n-1));
    }

    return 0;
}