Reward
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4638 Accepted Submission(s): 1416
Problem Description Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.
The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a's reward should more than b's.Dandelion's unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work's reward will be at least 888 , because it's a lucky number.
Input One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)
then m lines ,each line contains two integers a and b ,stands for a's reward should be more than b's.
Output For every case ,print the least money dandelion 's uncle needs to distribute .If it's impossible to fulfill all the works' demands ,print -1.
Sample Input
2 1
1 2
2 2
1 2
2 1
Sample Output
1777
-1
题意：有n个人，每个人最少发的奖金是888，现有m个要求，有m行a b，表示a 要求 b的奖金多。问能否满足所有的要求，如果能，则n个人得到的奖金总和最少是多少，不能满足所有要求就输出-1。
#include
#include
#include
#include
using namespace std; const int N = 10005; int n,mony[N],in[N]; vector
mapt[N]; void init() { for(int i=1;i<=n;i++) mony[i]=888,in[i]=0,mapt[i].clear(); } int tope() { queue
q; int k=0,sum=0; for(int i=1;i<=n;i++) if(in[i]==0) { k++; q.push(i); in[i]=-1; sum+=mony[i]; } while(!q.empty()) { int s=q.front(); q.pop(); for(int i=0;i
0) { in[j]--; if(mony[j]
0) { init(); while(m--) { scanf("%d%d",&a,&b); in[a]++; mapt[b].push_back(a); } printf("%d\n",tope()); } }
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