HDU-5943 Kingdom of Obsession（数学+二分图匹配）

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#二分图匹配 #数学

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本文探讨了一个关于整数编号的人们希望站在特定位置的问题，并通过数学分析与二分图匹配的方法给出了有效的解决方案。

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Kingdom of Obsession

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1202 Accepted Submission(s): 368

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

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

2
5 14
4 11

Sample Output

Case #1: No
Case #2: Yes

题解：数学+二分图匹配

打表可知：n和s存在对称关系，即n和s互换对答案没有影响
根据这一点，可以使n小于s。这样有什么用呢？

我们知道素数的约数只有1和本身，我们令n<s后，编号的区间为[s+1,s+n]

因此只要再区间[s+1,s+n]中有2个素数，那么这种情况肯定是No（因为[1,n]中只有一是这些素数的约数）

再根据10E以内相邻素数之差不会超过300，因此只要n>300即输出No

当n在300以内时，就可以用二分图匹配将编号和其约数配对，如果匹配成功则输出Yes

#include<bits/stdc++.h>
using namespace std;
const int MX = 2e3 + 5;
struct Edge{
    int v,nxt;
}E[MX*MX];
int head[MX],tot;
void init(){
    memset(head,-1,sizeof(head));
    tot=0;
}
void add(int u,int v){
    E[tot].v=v;
    E[tot].nxt=head[u];
    head[u]=tot++;
}
int match[MX];
bool vis[MX];
bool DFS(int u) {
    for(int i = head[u]; ~i; i = E[i].nxt) {
        int v = E[i].v;
        if(!vis[v]) {
            vis[v] = 1;
            if(match[v] == -1 || DFS(match[v])) {
                match[v] = u;
                return 1;
            }
        }
    }
    return 0;
}
int BM(int n) {
    int res = 0;
    memset(match, -1, sizeof(match));
    for(int u = 1; u <= n; u++) {
        memset(vis, 0, sizeof(vis));
        if(DFS(u)) res++;
    }
    return res;
}
int main(){
    int T,n,s;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++){
        printf("Case #%d: ",cas);
        scanf("%d%d",&n,&s);
        if(n>s) swap(n,s);
        if(n>600) {
            printf("No\n");
            continue;
        }
        init();
        for(int i=s+1;i<=s+n;i++){
            for(int j=1;j<=n;j++){
                if(i%j==0) {
                    add(i-s+n,j);
                    add(j,i-s+n);
                }
            }
        }
        if(BM(2*n)==2*n) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}