Hangzhou Electric 2647 Reward (topological sort reverse row)

DescriptionDandelion ' s uncle is a boss of a factory. As the Spring Festival is coming, he wants to distribute rewards to his workers. Now he had a trouble about what to distribute the rewards.
The workers would compare their rewards, and some one may has 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'll be at least 888, because it's a lucky number.InputOne line with a 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 the integers a and B, stands for a ' s reward should is more than B ' s.OutputFor 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 11 22 21 22 1

Sample Output

1777-1

Effect:
The farm mainly pay, but there are certain rules, some people have to pay more than some people, the minimum wage for everyone is 888. Enter two integer n,m, which represents the number of worker numbers 1~n and M rules, the next m line has two integers a, a, a is higher than B, enter the minimum amount of money, If there is no output-1.

Because there is no way to determine the wages of the highest wage is how much, so the reverse sort, so that the minimum wage in front, enter a A, B, the record of a constraint A, the predecessor of 0 of the salary of 888, followed by the cumulative.

1#include <cstdio>2#include <queue>3#include <string.h>4 using namespacestd;5 intn,m,i,j,num[20010],head[20010],mon[20010],sum,ans;6 structStu7 {8     intTo,next;9}st[20010];Ten voidInit () One { Amemset (NUM,0,sizeof(num)); -memset (head,-1,sizeof(head)); -memset (Mon,0,sizeof(Mon)); thesum=0; -ans=0; - } - voidAddintAintb) + { -st[i].to=b; +st[i].next=Head[a]; Ahead[a]=i; atnum[b]++; - } - voidtopo () - { -     inti,j; -queue<int>que; in      while(!que.empty ()) -     { to Que.pop (); +     } -      for(i =1; I <= N; i++) the     { *         if(Num[i] = =0) $         {Panax Notoginsengmon[i]=888; - Que.push (i); the         } +     } A      while(!que.empty ()) the     { +ans++; -m=Que.front (); $ Que.pop (); $sum+=Mon[m]; -          for(i = head[m]; I! =-1; i =st[i].next) -         { the             if(--num[st[i].to] = =0) -             {Wuyi Que.push (st[i].to); themon[st[i].to]=mon[m]+1; -             } Wu         } -     } About     if(ans = =N) $printf"%d\n", sum); -     Else -printf"-1\n"); - } A intMain () + { the     intb; -      while(SCANF ("%d%d", &n,&m)! =EOF) $     { the init (); the          for(i =0; I < m; i++) the         { thescanf"%d%d",&a,&b); - Add (b,a); in         } the topo (); the     } About      the}

Hangzhou Electric 2647 Reward (topological sort reverse row)