Kingdom of Obsession

標簽：?二分圖最大匹配 2016-10-29 16:23?51人閱讀?評論(2)?收藏?舉報 ?分類： 二分圖和最大匹配（2）?

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Kingdom of Obsession

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)

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 2 5 14 4 11
Sample Output Case #1: No Case #2: Yes

題意：我想來到這的都不用講什么題意的吧。

思路：估計也都能想到利用二分圖最大匹配來確定是不是能夠全匹配，重點是n很大啊。

不過我們知道素數有個特性就是因數只有1和它本身。所以素數只能待在位置1或者標號為本身的位置上

1.如果m>n

如果n大于等于兩個素數之間的距離，那么肯定是不可能匹配的。

如果小于直接二分圖匹配一下。

2.如果m<=n

黃色部分長度為m如果m的長度大于等于兩個素數的間隔那肯定是不符合題意。

如果小于那么就把m部分二分圖匹配，匹配的話肯定符合題意，不匹配的話不符合題意，至于為什么不和重合部分匹配，看評論。

這兩個一綜合就會發現第二種如果mn互換和第一種是一樣的。

至于二分圖匹配：

就是n個點之間的匹配，如果兩個點之間滿足等式就連一條邊，最后看看能不能全匹配上就可以了。

[cpp]?view plaincopy print?

#include?<iostream>??

#include?<cstdio>??

#include?<algorithm>??

#include?<cmath>??

#include?<queue>??

#include?<vector>??

#include?<cstring>??

#include?<string>??

using?namespace?std;??

const?int?MAXN=300+7;??

??

int?n,m;??

int?vis[MAXN],match[MAXN];??

int?num[MAXN];??

int?head[MAXN];??

int?cnt;??

struct?node??

{??

????int?v;??

????int?next;??

}?edge[MAXN*MAXN];??

??

void?add(int?u,int?v);??

int?dfs(int?u);??

int?build()??

{??

????cnt=0;??

????memset(match,-1,sizeof(match));??

????int?i,j,v;??

????for(i=1;?i<=n;?++i)head[i]=-1;??

????for(i=1;?i<=n;?++i)??

????{??

????????match[i]=-1;??

????????for(j=1;?j<=n;?++j)??

????????{??

????????????v=j+m;??

????????????if(v%i==0)add(i,j);??

????????}??

????}??

????int?sum=0;??

????for(i=1;?i<=n;?++i)??

????{??

????????memset(vis,0,sizeof(vis));??

????????if(dfs(i))sum++;??

????}??

????return?sum;??

}??

void?add(int?u,int?v)??

{??

????edge[cnt].v=v;??

????edge[cnt].next=head[u];??

????head[u]=cnt++;??

}??

int?dfs(int?u)??

{??

????int?i;??

????for(i=head[u];?i!=-1;?i=edge[i].next)??

????{??

????????int?v=edge[i].v;??

????????if(!vis[v])??

????????{??

????????????vis[v]=1;??

????????????if(match[v]==-1||dfs(match[v]))??

????????????{??

????????????????match[v]=u;??

????????????????return?1;??

????????????}??

????????}??

????}??

????return?0;??

}??

int?main()??

{??

????int?t;??

????scanf("%d",&t);??

????for(int?t1=1;?t1<=t;?++t1)??

????{??

????????scanf("%d%d",&n,&m);??

????????printf("Case?#%d:?",t1);??

????????if(m<n)swap(m,n);??

????????if(n>300)puts("No");??

????????else??

????????{??

????????????int?ans=build();??

????????????if(ans==n)puts("Yes");??

????????????else?puts("No");??

????????}??

??

????}??

????return?0;??

}??

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