HDU 2647 Reward （拓扑排序）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2647

题面：

Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6207    Accepted Submission(s): 1918


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

 

题目大意：
     老板发奖金，每人最少888元，有一些要求。a b，a的工资要比b的工资高。如果存在矛盾的关系，则输出-1，否则输出最好需要多少钱。

解题：
    比较明显的拓扑排序，与其他题目不同的是，节点较多，需用邻接表。同时需要给每个人分个权值，一个点可能会有多个点比它小，这时需要取所有它的权值中的最大值，确保它大于所有比它小的值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <queue>
using namespace std;
//s集合排重边，据说没有
set <int> s;
//分别为权值，入度，邻接表的头位置，cnt邻接表的计数值
int val[10010],in[10010],head[10010],cnt=0;
//访问标记
bool vis[10010];
//边存储
struct edge
{
int to,next;
}ee[20010];
//加边操作
void add_edge(int a,int b)
{
ee[cnt].to=b;
//接上后续地址
ee[cnt].next=head[a];
head[a]=cnt++;
}
queue <int> qe;
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{

int n,m,a,b,tmp,ans,cur,v,to;
bool flag;
while(~scanf("%d%d",&n,&m))
{
flag=true;
cnt=0;
ans=0;
while(!qe.empty())
qe.pop();
s.clear();
//赋初值
for(int i=1;i<=n;i++)
val[i]=888;
//初始化
memset(in,0,sizeof(in));
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
for(int i=0;i<m;i++)
{
//这写反了，wa了一次
scanf("%d%d",&b,&a);
tmp=a*10000+b;
//排重
if(!s.count(tmp))
{
//加边，统计入度，最小的点，入度为0
add_edge(a,b);
in[b]++;
}
}
for(int i=1;i<=n;i++)
{
if(in[i]==0)
{
qe.push(i);
ans+=888;
vis[i]=1;
}
}
while(!qe.empty())
{
cur=qe.front();
qe.pop();
v=val[cur]+1;
for(int i=head[cur];i!=-1;i=ee[i].next)
{
to=ee[i].to;
//取最大值，以确保满足
val[to]=max(val[to],v);
//不断加入入度为0的点
in[to]--;
if(in[to]==0)
{
ans+=val[to];
vis[to]=1;
qe.push(to);
}
}
}
//如果还有没访问过的点，就是存在矛盾
for(int i=1;i<=n;i++)
{
if(!vis[i])
{
flag=false;
break;
}
}
if(flag)
printf("%d\n",ans);
else
printf("-1\n");
}
return 0;
}