HDU 2647：Reward

Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8557    Accepted Submission(s): 2737

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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 11 22 21 22 1

 

Sample Output

1777-1
AC代码：
#include<iostream>#include<cstdio>#include<queue>#include<vector>#include<cstring>#include<stack>#define INF 0x3f3f3f3f#define MAXN 10003using namespace std;int n,m;         vector<int>node[MAXN];  ///容器来存放图int num[MAXN];          ///num[i]用来存放节点i的入度struct reward{    int index;          ///人的编号    int money;          ///它的钱数};int HaveTopSort()       ///判断是否存在环{    reward p,cur;    queue<reward>qu;    int ccount = 0;    int ans = 0;    for(int i = 1; i <= n; i++)    {        if(num[i]==0)       ///入度为0的点入队        {            p.index = i;            p.money = 888;            ans += p.money;            qu.push(p);            ccount++;       ///统计入队的点数        }    }    if(ccount == n) return ans;    while(!qu.empty())    {        cur = qu.front();        qu.pop();        for(int i = 0; i < node[cur.index].size(); i++)        {            int t = node[cur.index][i];            p.index = t;            p.money = cur.money+1;            num[t]--;              if(num[t]==0)              {                ans += p.money;                qu.push(p);                ccount++;                if(ccount == n)                    return ans;            }        }    }    return -1;}int main(){    while(~scanf("%d%d",&n,&m)) ///n个工人，m个要求    {        int a,b;        for(int i = 1; i <= n; i++)        {            node[i].clear();        ///清空容器            num[i] = 0;             ///节点i的入度初始化为0        }        for(int i = 0; i < m; i++)          {            scanf("%d%d",&a,&b);    ///a的奖金必须大于b的奖金。方向为b->a            node[b].push_back(a);            num[a]++;               ///入度加1，        }        int temp = HaveTopSort();        printf("%d\n",temp);    }    return 0;}