HDU 2647 Reward【拓扑排序】

Reward

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 44   Accepted Submission(s) : 24

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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

注意：用的广搜，所以只需要每次dis[i]=0时确定flag并入队列即可！！！（广搜情况下，不同reward相当于处于不同层次，只要dis不是零，

就说明再次遇到他时，已处于更高的层次）

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<climits>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
#define rep(i,j,k)for(i=j;i<k;i++)
#define per(i,j,k)for(i=j;i>k;i--)
#define MS(x,y)memset(x,y,sizeof(x))
typedef long long LL;
const int INF =0x7FFFFFFF;
const int low(int x){return x&-x;}

const int M=10000+1;
int dis[M];
int flag[M];
vector<int>s[M];
int i,j,n,m,k,t;

int Topo()
{
queue<int>Q;
rep(i,1,n+1)
if(!dis[i])
Q.push(i),flag[i]=888;
int num=0;
while(!Q.empty())
{
int cur=Q.front();
Q.pop();
num++;
rep(i,0,s[cur].size()){
dis[s[cur][i]]--;
if(!dis[s[cur][i]]){
Q.push(s[cur][i]);
flag[s[cur][i]]=flag[cur]+1;
}
}
}
if(num<n)return 0;
return 1;
}

int main()
{
while(~scanf("%d%d",&n,&m))
{
MS(flag,0);
MS(dis,0);
rep(i,0,n+1)s[i].clear();
rep(i,0,m){
int a,b;
scanf("%d%d",&a,&b);
s[b].push_back(a);
dis[a]++;
}
if(!Topo()){
printf("-1\n");
continue;
}
int sum=0;
rep(i,1,n+1)
sum+=flag[i];
printf("%d\n",sum);
}
return 0;
}