| Problem description |
|
Porto’s book club is buzzing with excitement for the annual book exchange event! Every year, members bring their favorite book and try to find another book they like that is owned by someone willing to trade with them. I have been to this book exchange before, and I definitely do not want to miss it this year, but I feel that the trading should be improved. In the past, pairs of members interested in each other’s books would simply trade: imagine that person A brought a book I then realized that many members were left with the same book they walked-in with... If instead of looking for pairs I looked for triplets, I could find more valid exchanges! Imagine that member A only likes member B’s book, while B only likes C’s book and But why stop at triplets? Cycles could be bigger and bigger! Could you help me find if it is possible for everyone to go out with a new book? Be careful, because members will not give their book without receiving one they like in return. Given the members of the book club and the books they like, can we find cycles so that everyone receives a new book? |
| Input |
|
The first line has two integers: N, the number of people, and M, the total number of “declarations of interest”. Each of the following M lines has two integers, A and B, indicating that member A likes the book that member B brought (0<=A,B < N). Numbers 1<=M<=20 000 and M<=N^2-N. |
| Output |
|
You should output YES if we can find a new book for every club member and NO if that is not possible. |
| Sample Input |
9 9 0 1 1 2 2 0 3 4 4 3 5 6 6 7 7 8 8 5 |
| Sample Output |
YES |
| Problem Source |
| HNU Contest |
题意：
有n个人，m种需求，给出m行，每行a,b代表a想要的书在b那里，问能不能通过交换的方法来满足每个人的需求
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 20005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define rank rank1
const int mod = 1000000007;
int n,m,vis[N],tem[N];
vector<int> a[N];
int dfs(int u)
{
for(int i=0; i<a[u].size(); i++)
{
int v = a[u][i];
if(!vis[v])
{
vis[v]=1;
if(tem[v]==-1||dfs(tem[v]))
{
tem[v] = u;
return 1;
}
}
}
return 0;
}
int main()
{
int i,j,k,x,y;
while(~scanf("%d%d",&n,&m))
{
for(i = 0; i<=n; i++)
a[i].clear();
for(i = 0; i<m; i++)
{
scanf("%d%d",&x,&y);
a[x].push_back(y);
}
MEM(tem,-1);
for(i = 0; i<n; i++)
{
MEM(vis,0);
dfs(i);
}
for(i = 0; i<n; i++)
{
if(tem[i]==-1)
break;
}
if(i==n)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
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