2016ccpc杭州赛 hdu 5934 K.Kingdom of Obsession

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本文探讨了一个关于整数编号人员按特定条件排列的问题，通过分析给出了一种解决方案，利用二分图匹配理论来判断是否能按照每个人希望的位置进行排列。

Kingdom of Obsession

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

Problem Description

There is a kindom of obsession, so people in this kingdom do things very strictly.

They name themselves in integer, and there are

n people with their id continuous

(s+1,s+2,⋯,s+n) standing in a line in arbitrary order, be more obsessively, people with id

x wants to stand at

yth position which satisfy

x mod y = 0

Is there any way to satisfy everyone's requirement?

Input

First line contains an integer

T , which indicates the number of test cases.

Every test case contains one line with two integers

n ,

s .

Limits

1≤T≤100 .

1≤n≤109 .

0≤s≤109 .

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the result string.

If there is any way to satisfy everyone's requirement, y equals 'Yes', otherwise y equals 'No'.

Sample Input

Sample Output

  
   Case #1: No
Case #2: Yes

Statistic | Submit | Clarifications | Back

怎么感觉这次杭州是图论专场，居然两题涉及图论。

题意：n个人，id序号分别为是s+1到s+n，他们各自想站在第i个位置（id%i==0），问全部人能否站在想要的位置上。

思路：一看就是二分图，但数据很大不可能一一建边，然后想到了只要id里存在两个素数就不可能匹配成功，然后大概估算一下每1000个肯定有两个素数吧。然后就是暴力n^2建边，跑一遍二分图就好了。这里有个问题，区间重叠，如果重叠的话就算存在两个素数也是可以匹配了，因为可以一一对应。所以这里要特殊处理一下，区间重叠的部分默认匹配成功，剩下的部分跑二分图就好了，一开始这里没想到，一直wa。。下面给代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<queue>
#include<utility>
#include<map>
#define maxn 2005
#define inf 0x3f3f3f3f
typedef long long LL;
using namespace std;
const int mod = 1e9 + 7;
int head[maxn], vis[maxn];
bool book[maxn];
struct node{
	int v, next;
}p[maxn*maxn];
bool find(int x)
{
	for (int i = head[x]; ~i; i = p[i].next)
	{
		int y = p[i].v;
		if (!book[y])//防止同一个人被搜两次 ,不然会死循环 
		{
			book[y] = true;
			if (!vis[y] || find(vis[y]))//没人占用或者占用你的人可以找到新欢，太好了 
			{
				vis[y] = x;
				return true;
			}
		}
	}
	return false;
}
int main(){
	int t;
	scanf("%d", &t);
	for (int tcase = 1; tcase <= t; tcase++){
		memset(head, -1, sizeof(head));
		memset(vis, 0, sizeof(vis));
		int n, s, len = 0;
		scanf("%d%d", &n, &s);
		if (s + 1 <= n)
			swap(n, s);
		if (n > 1000)
			printf("Case #%d: No\n", tcase);
		else{
			for (int i = s + 1; i <= s + n; i++){
				for (int j = 1; j <= n; j++){
					if (i%j)
						continue;
					int u = i - s + 1000;
					p[len].v = j;
					p[len].next = head[u];
					head[u] = len++;
					p[len].v = u;
					p[len].next = head[j];
					head[j] = len++;
				}
			}
			int ans = 0;
			for (int i = 1; i <= n; i++)
			{
				memset(book, false, sizeof(book));//用来标记配偶是否被搜过一次 
				if (find(i))ans++;//找到配偶啦！好开心。对数++ 
			}
			if (ans == n)
				printf("Case #%d: Yes\n", tcase);
			else
				printf("Case #%d: No\n", tcase);
		}
	}
}