Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13509    Accepted Submission(s): 4314

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 Recommend yifenfei   |   We have carefully selected several similar problems for you:  

3342 1811 2680 2112 2729  分析： a b：表示a的錢應該比b多，基本工資是888，所以a最少是889 坑點： 1.必須使用鄰接表存圖，鄰接矩陣會超記憶體 2.可能存在多條拓撲路徑，所以不能最後直接算總工資（有點沒有說明白，就是有的人工資可以和其他人一樣，看下面的兩組資料就知道了） 資料： 5 5
1 2
2 3
4 5
2 3
4 5

答案：4444

5 4
1 2
2 5
2 4
4 3


答案：4446 所以我們佇列裡面得存結構體！！！ code：

#include<stdio.h>
#include<iostream>
#include
 
<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
 
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};

int getval()
{
    int ret(0);
    char c;
    while((c=getchar())==' '||c=='\n'||c=='\r');
    ret=c-'0';
    while((c=getchar())!=' '&&c!='\n'&&c!='\r')
        ret=ret*10+c-'0';
    return ret;
}

#define max_v 10005
int indgree[max_v];
struct node
{
    int index,v;
    node(int x,int y)
    {
        index=x;
        v=y;
    }
};
vector <int> vv[max_v];
int n,m;
LL sum;
queue<node> q;
int tpsort()
{
    sum=0;
    while(!q.empty())
        q.pop();
    for(int i=1;i<=n;i++)
    {
        if(indgree[i]==0)
            q.push(node(i,888));
    }

    int c=0;
    int flag=0;
    int p1,p2;
    while(!q.empty())
    {
        if(q.size()>1)
            flag=1;
        p1=q.front().index;
        p2=q.front().v;
        q.pop();
        c++;
        sum+=p2;
        for(int i=0;i<vv[p1].size();i++)
        {
            indgree[vv[p1][i]]--;
            if(indgree[vv[p1][i]]==0)
                q.push(node(vv[p1][i],p2+1));
        }
    }
    /*
    拓撲完之後，存在沒有入隊的點，那麼該點就一定是環上的
    */
    if(c!=n)//存在環
        return 1;
    else if(flag)//能拓撲但存在多條路
        return 0;
    return -1;//能拓撲且存在唯一拓撲路徑
}
int main()
{
    int x,y;
    while(~scanf("%d %d",&n,&m))
    {
        memset(indgree,0,sizeof(indgree));
        for(int i=1;i<=n;i++)
        {
            vv[i].clear();
        }
        int flag=-1;
        for(int i=0;i<m;i++)
        {
            scanf("%d %d",&y,&x);
            if(count(vv[x].begin(),vv[x].end(),y)==0)//防止重邊
            {
                vv[x].push_back(y);
                indgree[y]++;
            }
            if(count(vv[y].begin(),vv[y].end(),x)!=0)//環的一種
            {
                flag=1;
            }
        }
        flag=tpsort();
        if(flag==1)
        {
            printf("-1\n");
        }else
        {
            printf("%lld\n",sum);
        }
    }
}