hdu 5943 Kingdom of Obsession（二分匹配）

Kingdom of Obsession

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1958    Accepted Submission(s): 581

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

xmody=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

25 144 11

 

Sample Output

Case #1: NoCase #2: Yes

 

Source

2016年中国大学生程序设计竞赛（杭州）

解：首先估计一下两个素数之间的距离大约是500~600，如果序列移动距离超过这个区间即一定存在两个素数

序列重合区间用被重合的数作为最终标号，因为将小的约数留给其他数一定更优

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int N = 600;typedef long long LL;int w[600][600], vis[600],match[N];int n, s;int dfs(int u){    for(int i=1;i<=n;i++)    {        if(!w[u][i]||vis[i]) continue;        vis[i]=1;        if(match[i]==-1||dfs(match[i]))        {            match[i]=u;            return 1;        }    }    return 0;}int main(){    int t, ncase=1;    scanf("%d", &t);    while(t--)    {        scanf("%d %d", &n, &s);        if(s<n)swap(s,n);        if(n>504) printf("Case #%d: No\n",ncase++);        else        {            memset(w,0,sizeof(w));            for(int i=1;i<=n;i++)            {                int x=i+s;                for(int j=1;j<=n;j++)                {                    if(x%j==0) w[i][j]=1;                }            }            int cnt=0;            memset(match,-1,sizeof(match));            for(int i=1;i<=n;i++)            {                memset(vis,0,sizeof(vis));                if(dfs(i)) cnt++;            }            if(cnt!=n) printf("Case #%d: No\n",ncase++);            else printf("Case #%d: Yes\n",ncase++);        }    }    return 0;}