hdu 2647 Reward (拓扑排序)
来源:互联网 发布:最浪漫的一句话知乎 编辑:程序博客网 时间:2024/03/28 16:15
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.
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.
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
拓扑排序+优先队列
#include"stdio.h"#include"string.h"#include"queue"using namespace std;#define N 10005#define M 20005struct 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<st>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;}
0 0
- HDU 2647 Reward(拓扑排序)
- hdu 2647 Reward (拓扑排序)
- hdu 2647 Reward ( 拓扑排序 )
- HDU 2647 - Reward(拓扑排序)
- hdu 2647 Reward (拓扑排序)
- HDU 2647Reward(拓扑排序)
- HDU 2647 Reward(图论-拓扑排序)
- HDU 2647 Reward(图论-拓扑排序)
- HDU 2647 Reward(拓扑排序)
- HDU 2647 -- Reward (拓扑排序)
- hdu 2647 Reward(拓扑排序)
- HDU-2647 Reward(拓扑排序)
- HDU 2647Reward (拓扑排序)
- HDU 2647 Reward(拓扑排序)
- HDU 2647 Reward(拓扑排序)
- HDU 2647 Reward(拓扑排序)
- HDU 2647 Reward (拓扑排序)
- HDU 2647 Reward(拓扑排序)
- 了解Linux文件系统_(2)
- 高内聚,低耦合?
- FLUSH TABLES WITH READ LOCK
- 结构体中冒号的作用
- 字符设备驱动分析
- hdu 2647 Reward (拓扑排序)
- response控制浏览器实现重定向
- ORA-12535:TNS连接超时/oracle客户端连接服务器报错解决办法
- 怎样把一个链表掉个顺序
- poj_1952最大下降子序列,统计个数
- linux--rsync--rsync安装、配置、实例
- 毕业到工作半年多的感受。
- VC中文文章汇总
- 在CentOS 6.4中支持exfat格式的U盘