题意：老板要给N个员工发工资,要求每个人最少888块且要满足M个要求.比如1号员工的工资要比2号员工的工资多.老板准备满足所有要求且最终发的奖金总数要最少.如果可行输出总数,不可行输出-1.
思路：就是将工资拓扑排序，注意的是输入的时候添加边要添加反向边。
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 50005
#define LL long long
int cas=1,T;
int n,m;
vector<int>G[maxn];
int in[maxn];
int money[maxn];
int sum;
bool topo()
{
queue<int>q;
for (int i = 1;i<=n;i++)
if (in[i]==0)
q.push(i);
sum = 0;
int ans = 0;
while (!q.empty())
{
int u = q.front();q.pop();
sum+=money[u];
ans++;
for (int i = 0;i<G[u].size();i++)
{
int v = G[u][i];
if (--in[v]==0)
{
q.push(v);
money[v]=money[u]+1;
}
}
}
return ans == n;
}
int main()
{
while (scanf("%d%d",&n,&m)!=EOF)
{
memset(in,0,sizeof(in));
memset(money,0,sizeof(money));
for (int i = 1;i<=n;i++)
G[i].clear();
for (int i = 1;i<=n;i++)
money[i]=888;
while (m--)
{
int a,b;
scanf("%d%d",&a,&b);
G[b].push_back(a);
in[a]++;
}
if (topo())
{
printf("%d\n",sum);
}
else
printf("-1\n");
}
//freopen("in","r",stdin);
//scanf("%d",&T);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
} 题目
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
