UVA 10330 Power Transmission

UVA_10030

对于题目中每个点的容量的限制，我们可以把一个点i拆成两个点2*i和2*i ^ 1，并连一条有向边2*i->2*i^1，容量为该点的容量限制，在建图的时候从2*i流入并从2*i^1流出即可。

#include<stdio.h>
#include<string.h>
#define MAXD 210
#define MAXM 81000
#define INF 100000000
int N, flow[MAXM], first[MAXD], next[MAXM], u[MAXM], v[MAXM], e;
int work[MAXD], d[MAXD], q[MAXD];
int add(int a, int b, int w)
{
    u[e] = a;
    v[e] = b;
    flow[e] = w;
    next[e] = first[a];
    first[a] = e;
    e ++;
}
int init()
{
    int i, j, a, b, w, M, B, D;
    if(scanf("%d", &N) != 1)
        return 0;
    e = 0;
    memset(first, -1, sizeof(first));
    for(i = 1; i <= N; i ++)
    {
        scanf("%d", &w);
        add(2 * i, 2 * i ^ 1, w);
        add(2 * i ^ 1, 2 * i, 0);
    }
    scanf("%d", &M);
    for(i = 0; i < M; i ++)
    {
        scanf("%d%d%d", &a, &b, &w);
        add(2 * a ^ 1, 2 * b, w);
        add(2 * b, 2 * a ^ 1, 0);
    }
    scanf("%d%d", &B, &D);
    for(i = 0; i < B; i ++)
    {
        scanf("%d", &j);
        add(0, 2 * j, INF);
        add(2 * j, 0, 0);
    }
    for(i = 0; i < D; i ++)
    {
        scanf("%d", &j);
        add(2 * j ^ 1, 1, INF);
        add(1, 2 * j ^ 1, 0);
    }
    return 1;
}
int bfs()
{
    int i, j, rear;
    memset(d, -1, sizeof(d));
    d[0] = 0;
    rear = 0;
    q[rear ++] = 0;
    for(i = 0; i < rear; i ++)
        for(j = first[q[i]]; j != -1; j = next[j])
            if(d[v[j]] == -1 && flow[j])
            {
                d[v[j]] = d[q[i]] + 1;
                if(v[j] == 1)
                    return 1;
                q[rear ++] = v[j];
            }
    return 0;
}
int dfs(int cur, int a)
{
    if(cur == 1)
        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], a < flow[i] ? a : flow[i]))
            {
                flow[i] -= t;
                flow[i ^ 1] += t;
                return t;
            }
    return 0;
}
int dinic()
{
    int res = 0, t;
    while(bfs())
    {
        memcpy(work, first, sizeof(first));
        while(t = dfs(0, INF))
            res += t;
    }
    return res;
}
int main()
{
    while(init())
    {
        int res = dinic();
        printf("%d\n", res);
    }
    return 0;
}