POJ 3498 March of the Penguins

POJ_3498

    对于任何一点来讲，限制的是跳走的企鹅的数量，如果这个点不是终点，那么必然跳过来的企鹅都会跳走，因此实际上限制的是经过这个点的企鹅的数量，这样通过拆点来限制经过点的企鹅的数量就可以了。最后暴力一点，枚举每个点作为终点并判断一下就可以了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXD 210
#define MAXM 40410
#define INF 0x3f3f3f3f
int N, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM];
int S, T, d[MAXD], q[MAXD], work[MAXD], list[MAXD], L, TOT;
double D;
struct Stone
{
    double x, y;
    int n, m;
}stone[MAXD];
double sqr(double x)
{
    return x * x;
}
void init()
{
    int i, j;
    TOT = 0;
    scanf("%d%lf", &N, &D);
    for(i = 0; i < N; i ++)
        scanf("%lf%lf%d%d", &stone[i].x, &stone[i].y, &stone[i].n, &stone[i].m), TOT += stone[i].n;
}
void add(int x, int y, int z)
{
    v[e] = y, flow[e] = z;
    next[e] = first[x], first[x] = e ++;
}
void build(int t)
{
    int i, j;
    S = 2 * N, T = t;
    memset(first, -1, sizeof(first[0]) * (N << 1 | 1)), e = 0;
    for(i = 0; i < N; i ++)
        add(i, i + N, stone[i].m), add(i + N, i, 0);
    for(i = 0; i < N; i ++)
        for(j = i + 1; j < N; j ++)
            if(sqr(D) >= sqr(stone[i].x - stone[j].x) + sqr(stone[i].y - stone[j].y))
                add(i + N, j, INF), add(j, i + N, 0), add(j + N, i, INF), add(i, j + N, 0);
    for(i = 0; i < N; i ++)
        if(stone[i].n)
            add(S, i, stone[i].n), add(i, S, 0);
}
int bfs()
{
    int i, j, rear = 0;
    memset(d, -1, sizeof(d[0]) * (N << 1 | 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)
{
    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])))
            {
                flow[i] -= t, flow[i ^ 1] += t;
                return t;
            }
    return 0;
}
int dinic()
{
    int ans = 0, t;
    while(bfs())
    {
        memcpy(work, first, sizeof(first[0]) * (N << 1 | 1));
        while(t = dfs(S, INF))
            ans += t;
    }
    return ans;
}
void solve()
{
    int i;
    L = 0;
    for(i = 0; i < N; i ++)
    {
        build(i);
        if(dinic() == TOT)
            list[L ++] = i;
    }
    if(L == 0)
        printf("-1\n");
    else
    {
        printf("%d", list[0]);
        for(i = 1; i < L; i ++) printf(" %d", list[i]);
        printf("\n");
    }
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        solve();
    }
    return 0;
}