Reward
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3554 Accepted Submission(s): 1077
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
拓扑排序+优先队列
#include"stdio.h"
#include"string.h"
#include"queue"
using namespace std;
#define N 10005
#define M 20005
struct st
{
int x,v; //存点坐标和价值
friend bool operator<(st a,st b)
{
return a.v>b.v;
}
};
struct node
{
int v,next;
}bian[M];
int indeg[N],e;
int head[N];
void add(int u,int v) //存边的信息
{
bian[e].v=v;
bian[e].next=head[u];
head[u]=e++;
}
int top_sort(int n)
{
int mark[N],i,x,cnt=0,sum=0,v;
memset(mark,0,sizeof(mark));
priority_queue
q;
st cur,next;
cur.v=888;
for(i=1;i<=n;i++)
{
if(indeg[i]==0)
{
cur.x=i;
mark[i]=1;
cnt++;
q.push(cur);
}
}
while(!q.empty()) //每次找出入度为零的点，更新它指向的点信息。
{
cur=q.top();
q.pop();
sum+=cur.v;
x=cur.x;
for(i=head[x];i!=-1;i=bian[i].next)
{
v=bian[i].v;
indeg[v]--;
if(indeg[v]==0&&!mark[v])
{
mark[v]=1;
cnt++;
next.v=cur.v+1;
next.x=v;
q.push(next);
}
}
}
if(cnt==n)
return sum;
return -1;
}
int main()
{
int n,m,i,v,u;
while(scanf("%d%d",&n,&m)!=-1)
{
memset(head,-1,sizeof(head));
memset(indeg,0,sizeof(indeg));
e=0;
while(m--)
{
scanf("%d%d",&v,&u);
add(u,v);
indeg[v]++;
}
int t=top_sort(n);
printf("%d\n",t);
}
return 0;
}