学步园

ZOJ_3229

    有源汇有上下界的最大流问题，具体的思路可以参考这篇博客：http://blog.csdn.net/water_glass/article/details/6823741，感觉按套路来做就可以了。注意最后输出的时候不是要输出所有的MM。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 375
#define MAXD 1375
#define MAXM 225740
#define INF 0x3f3f3f3f
int N, M, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM];
int S, T, SS, TT, d[MAXD], q[MAXD], work[MAXD], SUM;
struct List
{
    int n, id[110];
}list[MAXD];
int low[MAXN][1010];
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, j, x, G, D, C, high;
    S = 0, T = N + M + 1, SS = T + 1, TT = SS + 1;
    memset(first, -1, sizeof(first[0]) * (TT + 1)), e = 0;
    SUM = 0;
    for(i = 1; i <= M; i ++)
    {
        scanf("%d", &G);
        add(N + i, T, INF - G), add(T, N + i, 0);
        add(SS, T, G), add(T, SS, 0), add(N + i, TT, G), add(TT, N + i, 0);
        SUM += G;
    }
    for(i = 1; i <= N; i ++)
    {
        scanf("%d%d", &C, &D), list[i].n = C;
        add(S, i, D), add(i, S, 0);
        for(j = 0; j < C; j ++)
        {
            scanf("%d", &x), ++ x;
            list[i].id[j] = x;
            scanf("%d%d", &low[i][x], &high);
            add(i, N + x, high - low[i][x]), add(N + x, i, 0);
            add(SS, N + x, low[i][x]), add(N + x, SS, 0), add(i, TT, low[i][x]), add(TT, i, 0);
            SUM += low[i][x];
        }
    }
    add(T, S, INF), add(S, T, 0);
}
int bfs(int S, int T)
{
    int i, j, rear = 0;
    memset(d, -1, sizeof(d[0]) * (TT + 1));
    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, q[rear ++] = v[j];
                if(v[j] == T) return 1;
            }
    return 0;
}
int dfs(int cur, int a, int T)
{
    if(cur == T) return a;
    for(int &i = work[cur]; i != -1; i = next[i])
        if(flow[i] && d[v[i]] == d[cur] + 1)
            if(int t = dfs(v[i], std::min(a, flow[i]), T))
            {
                flow[i] -= t, flow[i ^ 1] += t;
                return t;
            }
    return 0;
}
int dinic(int S, int T)
{
    int ans = 0, t;
    while(bfs(S, T))
    {
        memcpy(work, first, sizeof(first[0]) * (TT + 1));
        while(t = dfs(S, INF, T))
            ans += t;
    }
    return ans;
}
void print()
{
    int i, j, k, ans = 0;
    for(i = first[S]; i != -1; i = next[i]) if(v[i] != T) ans += flow[i ^ 1];
    printf("%d\n", ans);
    for(i = 1; i <= N; i ++)
    {
        for(j = first[i]; j != -1; j = next[j])
            if(v[j] >= N + 1 && v[j] <= N + M)
                low[i][v[j] - N] += flow[j ^ 1];
        for(j = 0; j < list[i].n; j ++)
            printf("%d\n", low[i][list[i].id[j]]);
    }
}
void solve()
{
    if(SUM != dinic(SS, TT))
        printf("-1\n");
    else
    {
        dinic(S, T);
        print();
    }
}
int main()
{
    while(scanf("%d%d", &N, &M) == 2)
    {
        init();
        solve();
        printf("\n");
    }
    return 0;
}