HDU 2647 Reward(拓?fù)渑判颍?/h1>
來源：程序員人生 發(fā)布時(shí)間：2014-12-12 08:33:13 閱讀次數(shù)：3197次
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 ⑴.
Sample Input
Sample Output
Author
dandelion
拓?fù)渑判颉！?/p>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=10100;
int head[maxn*2],next[maxn*2],end[maxn*2];
int in[maxn],val[maxn];
int n,m,cnt,k;
void toposort()
{
int sum=0;
k=0;
queue<int>q;
REPF(i,1,n)
if(!in[i]) q.push(i);
while(!q.empty())
{
int v=q.front();
sum+=val[v];
q.pop();
k++;
for(int i=head[v];i!=⑴;i=next[i])
{
if(--in[end[i]]==0)
{
q.push(end[i]);
val[end[i]]=val[v]+1;
}
}
}
if(k==n) printf("%d
",sum);
else printf("⑴
");
}
void add(int x,int y)
{
next[cnt]=head[x];
end[cnt]=y;
head[x]=cnt++;
}
int main()
{
int x,y;
while(~scanf("%d%d",&n,&m))
{
cnt=0;
REPF(i,1,n) val[i]=888;
CLEAR(in,0);
CLEAR(head,⑴);
REP(i,m)
{
scanf("%d%d",&x,&y);
add(y,x);
in[x]++;
}
toposort();
}
return 0;
}
生活不易，碼農(nóng)辛苦
如果您覺得本網(wǎng)站對(duì)您的學(xué)習(xí)有所幫助,可以手機(jī)掃描二維碼進(jìn)行捐贈(zèng)
