2175: Leaders
Time Limit: 2 Sec Memory Limit: 256 MB
Submit: 26 Solved: 17
[Submit][Status][Web Board]
Description
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
After a revolution in Berland the new dictator faced an unexpected challenge: the country has to be somehow ruled. The dictator is a very efficient manager, yet he can't personally give orders to each and every citizen. That's why he decided to pick some set of leaders he would control. Those leaders will directly order the citizens. However, leadership efficiency turned out to vary from person to person (i.e. while person A makes an efficient leader, person B may not be that good at it). That's why the dictator asked world-famous berland scientists for help. The scientists suggested an innovatory technology − to make the leaders work in pairs.
A relationship graph is some undirected graph whose vertices correspond to people. A simple path is a path with no repeated vertices. Long and frighteningly expensive research showed that a pair of people has maximum leadership qualities if a graph of relationships has a simple path between them with an odd number of edges. The scientists decided to call such pairs of different people leader pairs. Secret services provided the scientists with the relationship graph so that the task is simple − we have to learn to tell the dictator whether the given pairs are leader pairs or not. Help the scientists cope with the task.
Input
The first line contains integers n and m (1≤n≤105,0≤m≤105) − the number of vertices and edges in the relationship graph correspondingly. Next m lines contain pairs of integers a and b which mean that there is an edge between the a-th and the b-th vertices (the vertices are numbered starting from 1, 1≤a,b≤n). It is guaranteed that the graph has no loops or multiple edges.
Next line contains number q (1≤q≤105) − the number of pairs the scientists are interested in. Next q lines contain these pairs (in the same format as the edges, the queries can be repeated, a query can contain a pair of the identical vertices).
Output
For each query print on a single line "Yes" if there's a simple odd path between the pair of people; otherwise, print "No".
Examples
Input
7 7
1 3
1 4
2 3
2 4
5 6
6 7
7 5
8
1 2
1 3
1 4
2 4
1 5
5 6
5 7
6 7
Output
No
Yes
Yes
Yes
No
Yes
Yes
Yes
Note
Notes to the samples:
1) Between vertices 1 and 2 there are 2 different simple paths in total: 1-3-2 and 1-4-2. Both of them consist of an even number of edges.
2) Vertices 1 and 3 are connected by an edge, that's why a simple odd path for them is 1-3.
【题意】
输入n，m。
n表示点的个数，再输入m行a，b分别表示边ab；
再输入q行x,y表示这两点之间的边是奇数还是偶数，奇数输出“Yes”，偶数输出“No”。
【解析】
原生无坑的建立无向图，求路径数，路径数为奇数为Yes，偶数为No。
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
vector<int> maps[N];
int vis[N];
int a, b, flag;
void dfs(int c, int d)//c是图顶点，d是边数
{
if (flag)return;//所有递归出路
if (c == b)//从a点能到b点
{
if (d % 2)flag = 1;//奇数时
return;//退出当前递归的栈，返回上一层
}
for (int i = 0; i < maps[c].size(); i++)//遍历所有该节点所有边
{
int n = maps[c][i];
if (vis[n])
{
vis[n] = 0;
dfs(n, d + 1);
vis[n] = 1;
}
}
}
int main()
{
int n, m, q;
while (~scanf("%d%d", &n, &m))
{
for (int i = 0; i < N; i++)
maps[i].clear();
for (int i = 0; i < m; i++)
{
scanf("%d%d", &a, &b);
maps[a].push_back(b);
maps[b].push_back(a);
}
scanf("%d", &q);
while (q--)
{
scanf("%d%d", &a, &b);
memset(vis, 1, sizeof(vis));
vis[a] = 0;
flag = 0;
dfs(a, 0);
if (flag)printf("Yes\n");
else printf("No\n");
}
}
return 0;
}