POJ 3687 Labeling Balls (反向拓扑排序)
来源:互联网 发布:淘宝开店 旺旺名字 编辑:程序博客网 时间:2024/05/16 23:42
Description
Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 to N in such a way that:
- No two balls share the same label.
- The labeling satisfies several constrains like "The ball labeled with a is lighter than the one labeled with b".
Can you help windy to find a solution?
Input
The first line of input is the number of test case. The first line of each test case contains two integers, N (1 ≤ N ≤ 200) and M (0 ≤ M ≤ 40,000). The next M line each contain two integers a and bindicating the ball labeled with a must be lighter than the one labeled with b. (1 ≤ a, b ≤ N) There is a blank line before each test case.
Output
For each test case output on a single line the balls' weights from label 1 to label N. If several solutions exist, you should output the one with the smallest weight for label 1, then with the smallest weight for label 2, then with the smallest weight for label 3 and so on... If no solution exists, output -1 instead.
Sample Input
54 04 11 14 21 22 14 12 14 13 2
Sample Output
1 2 3 4-1-12 1 3 41 3 2 4
题意:
使得小的结点尽可能的排在前面:需要 反向 拓扑排序
如:5->6->1; 4->3->2
应让结点1尽量排在前面,再使2尽量排在前面... ...:5 6 1 4 3 2
若直接按结点从小到大拓扑排序,无法得到正解: 4 3 2 5 6 1
反向拓扑排序,按结点从大到小的顺序这样能使结点小的尽可能的排在后面:2 3 4 1 6 5,那么反过来的5 6 1 4 3 2就是所求的拓扑序列,
已AC代码:
#include<cstdio>#include<cstring>int n,m;int map[300][300];int inde[41000];int que[300];void topo(){int i,k,top,t=0;for(k=n;k>=1;--k) //k 表示重量 从高到低依次选择可以选择的最大编号{top=-1;for(i=n;i>=1;--i) //j 表示每次可能选出的(即出度为0)最大编号,那么该编号一定能够排在其他所有编号的后边,{ //找出度为0的最大编号使其分配最大的重量,那么最后找出的解一定是最优解。if(inde[i]==0) {top=i;break;}}if(top==-1){printf("-1\n");return ;}//que[t++]=top;que[top]=k;inde[top]=-1;for(i=1;i<=n;++i){if(map[top][i]==1){inde[i]--;}}}printf("%d",que[1]);for(i=2;i<=n;++i)printf(" %d",que[i]);printf("\n");}int main(){int T,i,a,b;scanf("%d",&T);while(T--){memset(map,0,sizeof(map));memset(inde,0,sizeof(inde));scanf("%d%d",&n,&m);for(i=0;i<m;++i){scanf("%d%d",&a,&b);if(map[b][a]==0){map[b][a]=1;inde[a]++;}}topo();}return 0;}
- POJ 3687 Labeling Balls (反向拓扑排序)
- POJ 3687 Labeling Balls (反向拓扑排序)
- POJ 3687 Labeling Balls(拓扑排序+反向思考)
- poj 3687 Labeling Balls ( 拓扑排序 )
- **poj 3687 Labeling Balls (*拓扑排序)
- poj 3687 Labeling Balls(拓扑排序)
- poj 3687 Labeling Balls (拓扑排序)
- POJ 3687 Labeling Balls (拓扑排序)
- poj-3687-Labeling Balls-反向建图+拓扑排序
- Labeling Balls poj 3687(拓扑排序反向建图)
- POJ 3687 Labeling Balls 反向拓扑建图是个坑点
- 【POJ 3687 Labeling Balls】 + 反向拓扑
- poj 3687 Labeling Balls【反向拓扑】
- 3687-Labeling Balls-反向拓扑排序
- Labeling Balls 3687(拓扑排序+反向建图)
- poj 3687 Labeling Balls(拓扑排序)
- POJ 3687 Labeling Balls 拓扑排序
- poj 3687 【拓扑排序】【Labeling Balls】
- 树莓派2model B 通过蓝牙实现A2DP协议连接手机播放音乐
- cocos2dx 记录
- UITableView之上拉刷新
- 小心别让圆角成了你列表的帧数杀手 --关于设置圆角导致卡顿的深层原因解析
- DOS命令初接触
- POJ 3687 Labeling Balls (反向拓扑排序)
- 剑指offer_面试题26_复杂链表的复制
- 使用C++ 11 实现阻塞队列
- windows 消息机制的概述
- (一二三)基于GCD的dispatch_once实现单例设计
- 百股经项目部分知识点
- 关于C的未定义行为
- HDU 5386
- 操作系统的启动过程分析(以Linux系统为例)