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Problem H |
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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 by N (2 ≤ N ≤ 200). The road intersections are assumed to be numbered from 0 to N-1. The second number R denotes the number of roads (0 ≤ R ≤ 10000). Then there will be R lines each containing two numbers c1 and c2 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”.
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Sample Input |
Output for Sample Input |
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2 2 0 1 1 0 2 1 0 1 |
Possible Not Possible |
#include <iostream> #include <stack> #include <cstring> #include <cstdio> #include <string> #include <algorithm> #include <queue> #include <set> #include <map> #include <fstream> #include <stack> #include <list> #include <sstream> #include <cmath> using namespace std; #define ms(arr, val) memset(arr, val, sizeof(arr)) #define mc(dest, src) memcpy(dest, src, sizeof(src)) #define N 205 #define INF 0x3fffffff #define vint vector<int> #define setint set<int> #define mint map<int, int> #define lint list<int> #define sch stack<char> #define qch queue<char> #define sint stack<int> #define qint queue<int> int degree[N], fa[N], visit[N]; int n, r; int find(int x) { return fa[x] == -1 ? x : fa[x] = find(fa[x]); } void _union(int x, int y) { int fx = find(x), fy = find(y); if (fx != fy) { fa[fx] = fy; } } bool check() { int res = 0; for (int i = 0; i < n; i++) { if (visit[i] && fa[i] == -1) { res++; } } if (res != 1) { return false; } for (int i = 0; i < n; i++) { if (visit[i] && (degree[i] & 1)) { return false; } } return true; } void init() { ms(degree, 0); ms(fa, -1); ms(visit, 0); } int main() { int s, e; while (~scanf("%d", &n))//其实的无向图的欧拉回路，看网上说想了好久才想明白 { init(); scanf("%d", &r); while (r--) { scanf("%d%d", &s, &e); _union(s, e); degree[s]++; degree[e]++; visit[e] = visit[s] = 1;//可能存在孤立的点，所以需要标记 } if (check()) puts("Possible"); else puts("Not Possible"); } return 0; }
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原文：http://www.cnblogs.com/jecyhw/p/3914713.html