POJ 3687 拓扑排序应用
2016-02-09 23:00
267 查看
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 b indicating
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
Sample Output
#include <IOSTREAM>
#include <QUEUE>
#include <CSTDLIB>
using namespace std;
#define NUM 201
int gmap[NUM][NUM];
int N,M;
int degree[NUM];
int value[NUM];
priority_queue<int> que;
bool tuopo()
{
int remain=N;
int ic;
for (ic=1;ic<=N;ic++)
{
if (degree[ic]==0)
que.push(ic);
}
if (que.size()==0)
return false;
while (que.size()!=0)
{
int node=que.top();
value[node]=remain--;
que.pop();
for (ic=1;ic<=N;ic++)
{
if (gmap[ic][node]==1)
{
gmap[ic][node]=-1;
degree[ic]--;
if (degree[ic]==0)
que.push(ic);
}
}
}
if (remain==0)
return true;
else
return false;
}
int main()
{
int a,b,T;
cin>>T;
while (T!=0)
{
memset(gmap,0,sizeof(gmap));
memset(degree,0,sizeof(degree));
memset(value,0,sizeof(value));
N=M=0;
cin>>N>>M;
while (M!=0)
{
cin>>a>>b;
if(gmap[a][b]==0)
{
gmap[a][b]=1;
degree[a]++;
}
M--;
}
if (tuopo())
{
for (int ic=1;ic<=N;ic++)
{
cout<<value[ic]<<" ";
}
cout<<endl;
}
else
cout<<"-1"<<endl;
T--;
}
return 0;
}
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 b indicating
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
5 4 0 4 1 1 1 4 2 1 2 2 1 4 1 2 1 4 1 3 2
Sample Output
1 2 3 4 -1 -1 2 1 3 41 3 2 4
#include <IOSTREAM>
#include <QUEUE>
#include <CSTDLIB>
using namespace std;
#define NUM 201
int gmap[NUM][NUM];
int N,M;
int degree[NUM];
int value[NUM];
priority_queue<int> que;
bool tuopo()
{
int remain=N;
int ic;
for (ic=1;ic<=N;ic++)
{
if (degree[ic]==0)
que.push(ic);
}
if (que.size()==0)
return false;
while (que.size()!=0)
{
int node=que.top();
value[node]=remain--;
que.pop();
for (ic=1;ic<=N;ic++)
{
if (gmap[ic][node]==1)
{
gmap[ic][node]=-1;
degree[ic]--;
if (degree[ic]==0)
que.push(ic);
}
}
}
if (remain==0)
return true;
else
return false;
}
int main()
{
int a,b,T;
cin>>T;
while (T!=0)
{
memset(gmap,0,sizeof(gmap));
memset(degree,0,sizeof(degree));
memset(value,0,sizeof(value));
N=M=0;
cin>>N>>M;
while (M!=0)
{
cin>>a>>b;
if(gmap[a][b]==0)
{
gmap[a][b]=1;
degree[a]++;
}
M--;
}
if (tuopo())
{
for (int ic=1;ic<=N;ic++)
{
cout<<value[ic]<<" ";
}
cout<<endl;
}
else
cout<<"-1"<<endl;
T--;
}
return 0;
}
相关文章推荐
- Django 笔记 模型数据的读取
- Django 笔记 模型数据的读取
- Oracle学习第三天
- MySQL UPDATE子查询
- 什么叫做有能力?——做事果断、泼辣,效率高,独当一面,心思缜密。做人过得去就行
- js 表达式与运算符 详解(上)
- Why build CyanogenMod yourself?
- 乐视盒子UI 官方下载地址
- 算法导论第三版15-4整齐打印Printing neatly
- 乐视盒子新版系统ui3.0如何恢复成出厂系统ui1.5
- 一个人的浪漫
- ButterKnife的使用
- 《云计算核心技术剖析》学习笔记
- 使用nvm来管理nodejs版本
- u-boot-2016.01移植笔记之支持norflash
- 1011. World Cup Betting (20)
- #Android源码#MeasureSpec
- Kinect For Windows V2开发日志七:照片合成与背景消除
- 风雨阳光
- Cocos2d HttpClient 用法