UVA 10594 Data Flow

UVA_10594

    这个题目没看懂什么意思，但看样例觉得应该是把边的容量都设成K，把费用设成时间，然后求个从1到N的最小费用流。为了限制流量，我们可以连一条容量为D、费用为0的0->1这样的有向边，然后求图的最小费用最大流即可。

#include<stdio.h>
#include<string.h>
#define MAXD 110
#define MAXM 20100
const long long int INF = 10000000000000000ll;
int first[MAXD], next[MAXM], v[MAXM], a[MAXM], b[MAXM], N, M, e;
int p[MAXD], link[MAXD], q[MAXD], inq[MAXD];
long long int w[MAXM], cost[MAXM], flow[MAXM], D, K;
long long int d[MAXD];
void add(int x, int y, long long int c, long long int f)
{
    v[e] = y;
    cost[e] = c;
    flow[e] = f;
    next[e] = first[x];
    first[x] = e;
    e ++;
}
int init()
{
    int i;
    if(scanf("%d%d", &N, &M) != 2)
        return 0;
    for(i = 0; i < M; i ++)
        scanf("%d%d%lld", &a[i], &b[i], &w[i]);
    scanf("%lld%lld", &D, &K);
    e = 0;
    memset(first, -1, sizeof(first));
    for(i = 0; i < M; i ++)
    {
        add(a[i], b[i], w[i], K);
        add(b[i], a[i], -w[i], 0);
        add(b[i], a[i], w[i], K);
        add(a[i], b[i], -w[i], 0);
    }
    add(0, 1, 0, D);
    add(1, 0, 0, 0);
    return 1;
}
void SPFA()
{
    int i, u, front, rear;
    for(i = 1; i <= N; i ++)
        d[i] = INF;
    d[0] = 0;
    memset(inq, 0, sizeof(inq));
    front = rear = 0;
    q[rear ++] = 0;
    inq[0] = 1;
    while(front != rear)
    {
        u = q[front ++];
        if(front > N)
            front = 0;
        inq[u] = 0;
        for(i = first[u]; i != -1; i = next[i])
            if(flow[i] && d[u] + cost[i] < d[v[i]])
            {
                d[v[i]] = d[u] + cost[i];
                p[v[i]] = u;
                link[v[i]] = i;
                if(!inq[v[i]])
                {
                    q[rear ++] = v[i];
                    if(rear > N)
                        rear = 0;
                    inq[v[i]] = 1;
                }
            }
    }
}
long long int mincost()
{
    long long int res = 0, f = 0, a;
    int i, u;
    for(;;)
    {
        SPFA();
        if(d[N] == INF)
            break;
        a = INF;
        for(u = N; u != 0; u = p[u])
        {
            int y =link[u];
            if(flow[y] < a)
                a = flow[y];
        }
        for(u = N; u != 0; u = p[u])
        {
            int y = link[u];
            flow[y] -= a;
            flow[y ^ 1] += a;
        }
        res += a * d[N];
        f += a;
    }
    if(f == D)
        return res;
    else
        return -1;
}
int main()
{
    while(init())
    {
        long long int res = mincost();
        if(res >= 0)
            printf("%lld\n", res);
        else
            printf("Impossible.\n");
    }
    return 0;
}