HDU2647
2016-05-26 16:01
232 查看
Reward
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7031 Accepted Submission(s): 2187
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
Source
曾是惊鸿照影来
Recommend
yifenfei | We have carefully selected several similar problems for you: 3342 1811 2680 2112 2094
Statistic | Submit | Discuss | Note
思路:a b 表示a要比b多,所以会以b作为基础,那么久应该把b作为from(出度),a作为to(入度)。
#include <iostream> #include <cstdio> #include <queue> #include <vector> #include <cstring> using namespace std; const int maxn = 10005; int val[maxn]; int in[maxn]; vector <int > G[maxn]; int n,m,sum,ans; void topsort() { queue <int > que; while(!que.empty())que.pop(); for(int i = 1;i <= n;i++) { if(!in[i]) //这些人的奖金为888 que.push(i); } while(!que.empty()) { int u = que.front();que.pop(); sum += val[u]; ans++; for(int i = 0;i < G[u].size();i++) { int k = G[u][i]; if(--in[k] == 0) { val[k] = val[u] + 1; que.push(k); } } } } int main() { while(scanf("%d%d",&n,&m) != EOF) { for(int i = 0;i < maxn;i++) { val[i] = 888; G[i].clear(); } memset(in,0,sizeof(in)); sum = 0; ans = 0; int from,to; for(int i = 1;i <= m;i++) { scanf("%d%d",&to,&from); G[from].push_back(to); in[to]++; } topsort(); if(ans != n) sum = -1; printf("%d\n",sum); } return 0; }
#include <iostream> #include <cstdio> #include <queue> #include <cstring> using namespace std; const int maxn = 10001; int val[maxn],in[maxn],head[maxn]; int sum,ans,n,m; struct Node //链接表存边 { int to,next; }edge[maxn<<1]; void topsort() { queue <int >que; for(int i = 1;i <= n;i++) { if(in[i] == 0) //入度为0表示这个人奖金就是888 que.push(i); } while(!que.empty()) { int u = que.front();que.pop(); sum += val[u]; ans++; //用一个变量记录调用元素的总量,最后与n作比较 for(int i = head[u];i != -1;i = edge[i].next)//与队首元素u有关的都枚举一遍 { int k = edge[i].to; if(--in[k] == 0) //如果入度-1为0,即为u的下一个元素 { que.push(k); val[k] = val[u] + 1; } } } } int main() { while(scanf("%d%d",&n,&m) != EOF) { memset(head,-1,sizeof(head)); memset(in,0,sizeof(in)); for(int i=1; i<=n; i++) val[i]=888;//所有人一开始都为888 sum=0; ans=0; int from,to,tot=0; for(int i = 1;i <= m;i++) { scanf("%d%d",&to,&from); //注意要逆过来,因为后一个from是基础的888,应当作为出度 edge[tot].to = to; edge[tot].next = head[from]; head[from] = tot++; in[to]++; //入度 } topsort(); if(ans != n) sum = -1; printf("%d\n",sum); } return 0; }
相关文章推荐
- EXP-00003 ORA-01455 EXP导出报错
- java服务器端配置支持跨域请求
- iOSwebView修改字体大小字体颜色背景颜色
- Mac Pro下卸载安装Mysql
- arm 红外驱动
- form提交
- rigidbody.velocity 控制角色的移动
- LR11如何打开回放结果窗口 “Results.qtp”,
- 141. Linked List Cycle(Linked List)
- 模态对话框与非模态对话的几种销毁方法与区别
- this和super的使用
- php变量作用域的浅析
- ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2)
- C++Primer:Break、Continue、Goto、Try/异常处理
- 关于.so文件和处理器架构
- [Zookeeper系列一]Zookeeper应用介绍与安装部署
- Android Studio常用快捷键
- 网络与通信:网络编程(概念)
- winform ListView和DataGridView实现分页
- sqlserver数据以及日志文件的设置小结