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hdu 5116--Everlasting L(计数DP)

2018-06-17 22:20:27来源：未知阅读 ()

题目链接

Problem Description

Matt loves letter L.

A point set P is (a, b)-L if and only if there exists x, y satisfying:

P = {(x, y), (x + 1, y), . . . , (x + a, y), (x, y + 1), . . . , (x, y + b)}(a, b ≥ 1)

A point set Q is good if and only if Q is an (a, b)-L set and gcd(a, b) = 1.

Matt is given a point set S. Please help him find the number of ordered pairs of sets (A, B) such that:

Input

The first line contains only one integer T , which indicates the number of test cases.

For each test case, the first line contains an integer N (0 ≤ N ≤ 40000), indicating the size of the point set S.

Each of the following N lines contains two integers x_i, y_i, indicating the i-th point in S (1 ≤ x_i, y_i ≤ 200). It’s guaranteed that all (x_i, y_i) would be distinct.

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the number of pairs.

Sample Input

1 1

1 2

2 1

3 3

3 4

4 3

1 1

1 2

1 3

2 1

2 2

2 3

3 1

3 2

3 3

Sample Output

Case #1: 2

Case #2: 6

Hint

n the second sample, the ordered pairs of sets Matt can choose are:

A = {(1, 1), (1, 2), (1, 3), (2, 1)} and B = {(2, 2), (2, 3), (3, 2)}

A = {(2, 2), (2, 3), (3, 2)} and B = {(1, 1), (1, 2), (1, 3), (2, 1)}

A = {(1, 1), (1, 2), (2, 1), (3, 1)} and B = {(2, 2), (2, 3), (3, 2)}

A = {(2, 2), (2, 3), (3, 2)} and B = {(1, 1), (1, 2), (2, 1), (3, 1)}

A = {(1, 1), (1, 2), (2, 1)} and B = {(2, 2), (2, 3), (3, 2)}

A = {(2, 2), (2, 3), (3, 2)} and B = {(1, 1), (1, 2), (2, 1)}

Hence, the answer is 6.

题意：对于点集P 如果存在a,b使得P = {(x, y), (x + 1, y), . . . , (x + a, y), (x, y + 1), . . . , (x, y + b)}(a, b ≥ 1)，并且a，b互质，则P is good 。可以发现对于符合要求（good）的集合P ，其构成一个L 型，且以（x,y）为拐点，从（x,y）向上长度和向右长度互质。现在给了N个点，求有多少对符合要求的L型集合不相交（集合交集为空）？

思路：先找到所有符合要求的L个数S，那么用S*S-相交的L对数即为结果。

怎么算相交的所有L对数呢？容斥，很妙的思想，遍历每一个点，如果当前的点是输入的点之一，那么是一个拐点，令这个拐点向右延伸最长为k，那么算出所有其它L的竖着部分与（x,y+k）相交的对数，乘以2，另外要考虑以（x,y）为拐点的L与自身相交的情况，把这两种相交情形减掉后既是结果。

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int N=40050;
const int M=205;
int R[M][M],U[M][M];
bool mp[M][M];
int dp[M][M],cnt[M][M];
int t[M][M];

int gcd(int a,int b) { return (b==0)?a:gcd(b,a%b); }

void init()
{
    for(int i=1;i<M;i++)
    for(int j=1;j<M;j++)
    {
        dp[i][j]=dp[i][j-1]+((gcd(i,j)==1)?1:0);
        cnt[i][j]=cnt[i-1][j]+dp[i][j];
    }
}
int main()
{
    init();
    int T,Case=1;
    cin>>T;
    while(T--)
    {
        int n;  scanf("%d",&n);
        memset(mp,0,sizeof(mp));
        memset(U,0,sizeof(U));
        memset(R,0,sizeof(R));
        memset(t,0,sizeof(t));
        for(int i=1;i<=n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            mp[x][y]=1;
        }
        for(int i=200;i>=1;i--)
        {
            for(int j=200;j>=1;j--)
            {
                if(mp[i][j]){
                   if(mp[i+1][j]) U[i][j]=U[i+1][j]+1;
                   if(mp[i][j+1]) R[i][j]=R[i][j+1]+1;
                }
            }
        }
        LL s=0;
        for(int i=1;i<=200;i++)
        {
            for(int j=1;j<=200;j++)
            {
                if(mp[i][j]){
                    s+=cnt[U[i][j]][R[i][j]];
                    int d=0;
                    for(int k=U[i][j];k>=0;k--)
                    {
                        d+=dp[k][R[i][j]];
                        t[i+k][j]+=d;
                    }
                }
            }
        }
        LL ans=0;
        for(int i=1;i<=200;i++)
        {
            for(int j=1;j<=200;j++)
            {
                if(mp[i][j]){
                    LL p=t[i][j];
                    LL pp=cnt[U[i][j]][R[i][j]];
                    p-=pp;
                    for(int k=1;k<=R[i][j];k++)
                    {
                        p+=t[i][j+k];
                        ans+=2*p*dp[k][U[i][j]];
                    }
                    ans+=pp*pp;
                }
            }
        }
        s=s*s-ans;
        printf("Case #%d: %lld\n",Case++,s);
    }
    return 0;
}