2018山东省赛F-Four-tuples | 容斥定理

F-Four-tuples

题目描述

Given $l_1$ , $r_1$ , $l_2$ , $r_2$ , $l_3$ , $r_3$ , $l_4$ , $r_4$ ,please count the number of four-tuples ( $x_1$ , $x_2$ , $x_3$ , $x_4$ ) such that $l_i$ $≤$ $x_i$ $≤$ $r_i$ and $x_1$ $≠$ $x_2$ , $x_2$ $≠$ $x_3$ , $x_3$ $≠$ $x_4$ , $x_4$ $≠$ $x_1$ .The answer should modulo109+7 before output.

输入描述:

The input consists of several test cases. The first line gives the number of test cases,T(1≤T≤ $10^6$ ).
For each test case, the input contains one line with 8 integers $l_1$ , $r_1$ , $l_2$ , $r_2$ , $l_3$ , $r_3$ , $l_4$ , $r_4$ .(1 $≤$ $l_i$ $≤$ $r_i$ $≤$ $10^9$ )

输出描述:

For each test case, output one line containing one integer, representing the answer.

示例1

输入
1
1 1 2 2 3 3 4 4
输出
1

题解：

首先看这个题，容易想到是容斥。设事件A为 $x_1$ $≠$ $x_2$ ,事件B为 $x_2$ $≠$ $x_3$ ,事件C为 $x_3$ $≠$ $x_4$ ,事件D为 $x_4$ $≠$ $x_1$
问题要求： $|A∩B∩C∩D|$ ,可以转换成 $|U|-|A'∪B'∪C'∪D'|$ ,即求 $|A'∪B'∪C'∪D'|$ 就行，容斥定理:
这里写图片描述
因为相等的传递性， $x_1$ = $x_2$ , $x_2$ = $x_3$ , $x_3$ = $x_4$ 和 $x_1$ = $x_2$ , $x_2$ = $x_3$ , $x_4$ = $x_1$ 和 $x_1$ = $x_2$ , $x_3$ = $x_4$ , $x_4$ = $x_1$ 和 $x_2$ = $x_3$ , $x_3$ = $x_4$ , $x_4$ = $x_1$ 和 $x_1$ = $x_2$ , $x_2$ = $x_3$ , $x_3$ = $x_4$ , $x_4$ = $x_1$ 是一样的结果,所以只要减去三倍 $|A∩B∩C|$ 就可以了。

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define LL long long

using namespace std;

const int maxn=4;
const int mod=1e9+7;
LL L[maxn],R[maxn];

LL work2(int a,int b) //a==b
{
    LL mL=max(L[a],L[b]);
    LL mR=min(R[a],R[b]);
    if(mR<mL) return 0;
    return (mR-mL+1)%mod;
}

LL work3(int a,int b,int c) //a==b,c==d,e>=0
{
    LL mL1=max(L[a],max(L[b],L[c]));
    LL mR1=min(R[a],min(R[b],R[c]));
    if(mL1>mR1) return 0;
    return (mR1-mL1+1)%mod;
}

LL work4(int a,int b,int c,int d)
{
    LL cnt=1;
    LL mL=max(L[d],max(L[a],max(L[b],L[c])));
    LL mR=min(R[d],min(R[a],min(R[b],R[c])));

    if(mL>mR) return 0;
    return (mR-mL+1)*3%mod;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        for(int i=0; i<maxn; ++i)
            scanf("%lld %lld",&L[i],&R[i]);
        LL ans=1;
        for(int i=0; i<maxn; ++i)
        {
            ans=(ans*(R[i]-L[i]+1))%mod;
        }

        ans=(ans-work2(0,1)*(R[2]-L[2]+1)%mod*(R[3]-L[3]+1)%mod+mod)%mod;//0=1
        ans=(ans-work2(1,2)*(R[0]-L[0]+1)%mod*(R[3]-L[3]+1)%mod+mod)%mod;//1=2
        ans=(ans-work2(2,3)*(R[0]-L[0]+1)%mod*(R[1]-L[1]+1)%mod+mod)%mod;//2=3
        ans=(ans-work2(3,0)*(R[2]-L[2]+1)%mod*(R[1]-L[1]+1)%mod+mod)%mod;//3=0

        ans=(ans+work3(0,1,2)*(R[3]-L[3]+1)%mod)%mod;//0==1,1==2
        ans=(ans+work2(0,1)*work2(2,3)%mod)%mod;//0==1,2==3
        ans=(ans+work3(0,1,3)*(R[2]-L[2]+1)%mod)%mod;//0==1,3==0
        ans=(ans+work3(1,2,3)*(R[0]-L[0]+1)%mod)%mod;//1==2,2==3
        ans=(ans+work2(1,2)*work2(3,0)%mod)%mod;//1==2,3==0
        ans=(ans+work3(2,3,0)*(R[1]-L[1]+1)%mod)%mod;//2==3,3==0

        ans=(ans-work4(0,1,2,3)+mod)%mod;//0==1,1==2,2==3 ->3==0
        printf("%lld\n",ans);
    }
    return 0;
}