Description
Problem H
Morning Walk
Time Limit
3 Seconds
Kamal is a Motashota guy. He has got a new job in Chittagong. So, he has moved to Chittagong from Dinajpur. He was getting fatter in Dinajpur as he had no work in his hand there. So, moving to Chittagong has turned to be a blessing for him. Every morning he takes a walk through the hilly roads of charming city Chittagong. He is enjoying this city very much. There are so many roads in Chittagong and every morning he takes different paths for his walking. But while choosing a path he makes sure he does not visit a road twice not even in his way back home. An intersection point of a road is not considered as the part of the road. In a sunny morning, he was thinking about how it would be if he could visit all the roads of the city in a single walk. Your task is to help Kamal in determining whether it is possible for him or not.
Input
Input will consist of several test cases. Each test case will start with a line containing two numbers. The first number indicates the number of road intersections and is denoted byN (2 ≤ N ≤ 200). The road intersections are assumed to be numbered from0 to N-1. The second numberR denotes the number of roads (0 ≤ R ≤ 10000). Then there will beR lines each containing two numbersc1 andc2 indicating the intersections connecting a road.
Output
Print a single line containing the text “Possible” without quotes if it is possible for Kamal to visit all the roads exactly once in a single walk otherwise print “Not Possible”.
Sample Input
Output for Sample Input
2 2
0 1
1 0
2 1
0 1
Possible
Not Possible
Problemsetter: Muhammad Abul Hasan
International Islamic University Chittagong
需要注意一點：邊不是有向邊，兩點間可以有多條無向邊。
下面是錯誤且AC代碼。。。多連通分支情況不行。。。
算法：判斷圖是否連通且含歐拉回路就好，所以只要記錄點度數是否為零，且是否為偶數即可。當且僅當圖連通且所有點度數為偶數，輸出Possible :).
[cpp]
#include<iostream>
#include<cstdio>
#include<cstring>
#define N 202
using namespace std;
int u[N];
int main(){
int r,n;
while(cin>>n>>r){
memset(u,0,sizeof(u));
for(int i=0;i<r;i++){
int t1,t2;
cin>>t1>>t2;
u[t1]++;
u[t2]++;
}
int i;
for(i=0;i<n;i++){
if(!u[i] || (u[i]&1)) break;
}
if(i==n)cout<<"Possible"<<endl;
else cout<<"Not Possible"<<endl;
}
return 0;
}
#include<iostream>
#include<cstdio>
#include<cstring>
#define N 202
using namespace std;
int u[N];
int main(){
int r,n;
while(cin>>n>>r){
memset(u,0,sizeof(u));
for(int i=0;i<r;i++){
int t1,t2;
cin>>t1>>t2;
u[t1]++;
u[t2]++;
}
int i;
for(i=0;i<n;i++){
if(!u[i] || (u[i]&1)) break;
}
if(i==n)cout<<"Possible"<<endl;
else cout<<"Not Possible"<<endl;
}
return 0;
}
