reward

关键点是每一次点度数减1后，要即时更新其奖金值。。 用邻接矩阵写超时，学了下用STL vector写邻接表。用stl真的非常方便但是非常要时间啊

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <stack>

using namespace std;

int N, M, i, j ,t1, t2, sum ,l, a, b;
int dp[10010],exp[10010];

void init( )
{
    
    for (i = 0; i <= 10000; i++)
        exp[i] = 888;

}

int max(int x, int y)
{
   return x > y ? x : y;
}

int main( )
{
    
    while (scanf("%d%d", &N, &M)!= EOF) {
        init( );
        memset(dp, 0, sizeof(dp));
        sum = 0, l = 0;
        stack<int>q;
        vector<int>T[10010];
        for (i = 1; i <= M; i++) {
            scanf("%d%d", &a, &b);
            T[b].push_back(a); 
            dp[a]++;
        }
        for (i = 1; i <= N; i++)
            if (dp[i] == 0) {
                q.push(i);
                sum += exp[i];
               // printf("sum :%d exp[%d] : %d\n",sum,i,exp[i]); 
                dp[i] = -1;
                l++;
            }
            
        if (l == 0) {
            printf("-1\n");
            continue;
        }
        //puts("**");
        while (!q.empty()) {
            t1 = q.top();
            q.pop( );
            for (j = 0; j < T[t1].size( ); j++) {
                t2 = T[t1][j];
                dp[t2]--;
                exp[t2] = max(exp[t1]+1,exp[t2]);
            }
            for (i = 1; i <= N; i++)
                if (dp[i] == 0) {
                    q.push(i);
                    sum += exp[i];
                   // printf("sum :%d exp[%d] : %d\n",sum,i,exp[i]);
                    dp[i] = -1;
                    l++;
                }
            }
            //printf("%d\n",l);
            if(l == N)
                printf("%d\n",sum);
            else
                printf("-1\n");
   }
   return 0;
}