URAL 1664 Pipeline Transportation

URAL 1664

这个题目中间有一大段是用来迷惑的，其实源点就是1，汇点就是N。

最后输出的时候注意流量要从A流到B即可，并且每条边要顺序输出。

#include<stdio.h>
#include<string.h>
#define INF 0x7FFFFFFF
#define MAXD 10010
#define MAXM 10000010
int first[MAXD], next[MAXM], u[MAXM], v[MAXM], N, M, e;
long long int cap[MAXM], flow[MAXM];
int S, T, work[MAXD], d[MAXD], q[MAXD], s[MAXD];
void add(int a, int b, int w)
{
    u[e] = a;
    v[e] = b;
    cap[e] = flow[e] = w;
    next[e] = first[a];
    first[a] = e;
    e ++;
}
int init()
{
    int i, minx, maxx, x, y, a, b, w;
    if(scanf("%d", &N) != 1)
        return 0;
    minx = INF;
    maxx = 0;
    for(i = 1; i <= N; i ++)
        scanf("%d%d", &x, &y);
    S = 1;
    T = N;
    scanf("%d", &M);
    memset(first, -1, sizeof(first));
    e = 0;
    for(i = 0; i < M; i ++)
    {
        scanf("%d%d%d", &a, &b, &w);
        add(a, b, w);
        add(b, a, w);
    }
    return 1;
}
int bfs()
{
    int i, j, rear;
    memset(d, -1, sizeof(d));
    rear = 0;
    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 dinic()
{
    int cur, res = 0, r;
    while(bfs())
    {
        memcpy(work, first, sizeof(first));
        r = 0;
        cur = S;
        for(;;)
        {
            if(cur == T)
            {
                int minr = 0, a = INF;
                for(int i = 0; i < r; i ++)
                    if(flow[s[i]] < a)
                    {
                        a = flow[s[i]];
                        minr = i;
                    }
                for(int i = 0; i < r; i ++)
                {
                    flow[s[i]] -= a;
                    flow[s[i] ^ 1] += a;
                }
                r = minr;
                cur = u[s[r]];
                res += a;
            }
            int i;
            for(i = work[cur]; i != -1; i = next[i])
            {
                if(flow[i] == 0)
                    continue;
                if(d[cur] + 1 == d[v[i]])
                    break;
            }
            work[cur] = i;
            if(i != -1)
            {
                s[r ++] = i;
                cur = v[i];
            }
            else
            {
                d[cur] = -1;
                if(r == 0)
                    break;
                r --;
                cur = u[s[r]];
            }
        }
    }
    return res;
}
void printresult()
{
    int i;
    for(i = 0; i < 2 * M; i += 2)
    {
        if(flow[i] > cap[i])
            printf("%d %d %I64d\n", u[i ^ 1], v[i ^ 1], cap[i ^ 1] - flow[i ^ 1]);
        else
            printf("%d %d %I64d\n", u[i], v[i], cap[i] - flow[i]);
    }
}
int main()
{
    while(init())
    {
        long long int res = dinic();
        printf("%I64d\n", res);
        printresult();
    }
    return 0;
}