Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1803    Accepted Submission(s): 525

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  

Author dandelion  

Source
曾是惊鸿照影来  

Recommend yifenfei

题目本身没有什么好多说的，因为数据范围很大，用邻接矩阵做空间不够，所以用邻接表 第一次用vector做邻接表，还是记录一下

#include <stdio.h>
#include <string.h>
#include <vector>
using namespace std;

int n;
int indegree[10005];
vector <int>  v[10005];

int topsort(){
    int i,j,tag;
    int countn = 0,q = 0;
    int tem[10005];
    int sum = n * 888;

    while(countn != n){
        tag = 0;
        for(i = 1;i <= n;i++){
            if(!indegree[i]){
                indegree[i] = -1;
                tem[tag] = i;
                tag++;
            }
        }
        if(!tag){
            return -1;
        }
        else{
            sum += q * tag;
            q++;
            countn += tag;
            for(i = 0;i < tag;i++){
                for(j = 0;j < v[tem[i]].size();j++){
                    indegree[v[tem[i]].at(j)]--;
                }
            }
        }
    }
    return sum;
}
int main(){
    int m;
    int i,j;
    int a,b;

    while(scanf("%d%d",&n,&m) != EOF){
        for(i = 0;i < 10005;i++)
            v[i].clear();
        memset(indegree,0,sizeof(indegree));
        while(m--){
            scanf("%d%d",&a,&b);
            int flag = 0;
            for(i = 0;i < v[b].size();i++){
                if(v[b].at(i) == a)
                    flag = 1;
            }
            if(!flag){
                indegree[a]++;
                v[b].push_back(a);
            }
        }
        printf("%d\n",topsort());
    }
    return 0;
}