POJ 1637 Sightseeing tour

POJ_1637

    前些日子被各种网络流狂虐一番，现在准备好好搞一搞网络流的题目了，这题果断又被虐了，推荐一篇Edelweiss的网络流建模总结：http://wenku.baidu.com/view/0ad00abec77da26925c5b01c.html。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXD 210
#define MAXM 2410
#define INF 0x3f3f3f3f
int S, T, N, M, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM];
int indgr[MAXD], outdgr[MAXD], work[MAXD], q[MAXD], d[MAXD];
void add(int x, int y, int z)
{
    v[e] = y, flow[e] = z;
    next[e] = first[x], first[x] = e ++;    
}
void init()
{
    int i, x, y, k;
    scanf("%d%d", &N, &M);
    memset(first, -1, sizeof(first));
    e = 0;
    memset(indgr, 0, sizeof(indgr));
    memset(outdgr, 0, sizeof(outdgr));
    for(i = 0; i < M; i ++)
    {
        scanf("%d%d%d", &x, &y, &k);
        ++ outdgr[x], ++ indgr[y];
        if(k == 0)
            add(x, y, 1), add(y, x, 0);
    }
}
int bfs()
{
    int i, j, k, rear = 0;
    memset(d, -1, sizeof(d[0]) * (N + 2));
    d[S] = 0, q[rear ++] = S;
    for(i = 0; i < rear; i ++)
        for(j = first[q[i]]; j != -1; j = next[j])
            if(flow[j] && d[v[j]] == -1)
            {
                d[v[j]] = d[q[i]] + 1;
                if(v[j] == T)
                    return 1;
                q[rear ++] = v[j];    
            }
    return 0;
}
int dfs(int cur, int a)
{
    if(cur == T)
        return a;
    int t;
    for(int &i = work[cur]; i != -1; i = next[i])
        if(flow[i] && d[v[i]] == d[cur] + 1)
            if(t = dfs(v[i], a < flow[i] ? a : flow[i]))
            {
                flow[i] -= t, flow[i ^ 1] += t;
                return t;    
            }
    return 0;
}
int dinic()
{
    int t, ans = 0;    
    while(bfs())
    {
        memcpy(work, first, sizeof(first[0]) * (N + 2));
        while(t = dfs(S, INF))
            ans += t;
    }
    return ans;
}
void solve()
{
    int i, j, k, tot = 0;
    S = 0, T = N + 1;
    for(i = 1; i <= N; i ++)
    {
        k = abs(indgr[i] - outdgr[i]);
        if(k & 1)
        {
            printf("impossible\n");
            return ;    
        }
        k >>= 1;
        if(indgr[i] > outdgr[i])
            add(i, T, k), add(T, i, 0);
        else if(indgr[i] < outdgr[i])
            add(S, i, k), add(T, i, 0);
        tot += k;
    }
    if(dinic() == tot >> 1)
        printf("possible\n");
    else
        printf("impossible\n");
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        solve();    
    }
    return 0;    
}