【HDOJ】4857 逃生

很容易想到优先队列+拓扑排序。关键点是有限制条件者有限，无限制条件者在最后，条件相同者按序输出。因此采用逆序。

#include <iostream>
#include <queue>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;

#define MAXN 30005

typedef struct {
    int to, next;
} edge_st;

edge_st edges[100005];
int adj[MAXN], L;
int deg[MAXN];
int buf[MAXN];

typedef struct node_st{
    int i;
    node_st() {}
    node_st(int ii) {
        i = ii;
    }
    friend bool operator <(node_st a, node_st b) {
        return a.i < b.i;
    }
} node_st;

void add(int u, int v) {
    edges[L].to = v;
    edges[L].next = adj[u];
    adj[u] = L++;
    ++deg[v];
}

void init() {
    L = 1;
    memset(deg, 0, sizeof(deg));
    memset(adj, 0, sizeof(adj));
}

int main() {
    int T, n, m;
    int i, j, k;
    node_st nd;

    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &n, &m);
        init();
        while (m--) {
            scanf("%d %d", &j, &k);
            add(k, j);
        }
        priority_queue<node_st> Q;
        for (i=1; i<=n; ++i)
            if (deg[i] == 0)
                Q.push(node_st(i));
        k = 0;
        while (!Q.empty()) {
            nd = Q.top();
            buf[k++] = nd.i;
            Q.pop();
            for (i=adj[nd.i]; i; i=edges[i].next) {
                deg[edges[i].to]--;
                if (deg[edges[i].to] == 0)
                    Q.push(node_st(edges[i].to));
            }
        }
        for (i=k-1; i>=0; --i)
            if (i)
                printf("%d ", buf[i]);
            else
                printf("%d
", buf[i]);
    }

    return 0;
}