湘潭1247 Pair-Pair(树状数组)

Posted 2020-08-09

分析:

给定n个二元组，求选出两个二元组(可以是同一个)组成一序列其LIS为1,2,3,4的方法数。

分别记为s1, s2, s3, s4

 

s1,s4对应的情形为a >= b >= c >= d, a < b < c < d，易求

 

长度为3时，先求得s3 + s4的值，分解为两种情况的和减去两种情况的并，min(a, b) < c < d, a < b < max(c, d)，减去a < min(b, c) <= max(b, c) < d的方法数(使用二位树状数组，只考虑x[i] < y[i]),此时方法数为s3 + s4,减去s4得s3 

 

总数为n * n,减去其他情况即为s2

 

若有更好的解法请指出！

 

 

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 100008;
int C[1018];
int x[N], y[N];

inline int lowbit(int x){
    return x&-x;
}
void add(int x, int n){//将第x个数增加val，从1计数
    for(int i=x;i<=n;i+=lowbit(i)){
        C[i]++;
    }
}
int sum(int x){//求1到x的和
    int ret = 0;
    for(int i=x;i>0;i-=lowbit(i)){
        ret+=C[i];
    }
    return ret;
}
namespace bit{

int C[1008][1008];
inline int lowbit(int x){
    return x&-x;
}
void add(int x,int y,int n){
    for(int i=x;i<=n;i+=lowbit(i)){
        for(int j=y;j<=n;j+=lowbit(j)) {
            C[i][j]++;
        }
    }
}

int sum(int x,int y){
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i)) {
        for(int j=y;j>0;j-=lowbit(j)) {
            ret+=C[i][j];
        }
    }
    return ret;
}
	LL solve(int n){
		LL ans = 0;
		for(int i = 1; i <= n; i++){
			if(x[i] < y[i]){
				ans += sum(x[i] - 1, y[i] - 1);
			}
		}
		return ans;
	}

}

int main(){
	int n, m;
	while(~scanf("%d %d", &n, &m)){
        memset(bit::C, 0, sizeof(bit::C));
        int tot = 0;
		for(int i = 1; i <= n; i++){
			scanf("%d %d", &x[i], &y[i]);
			if(x[i] < y[i]){
                bit::add(x[i], y[i], m);
			}
		}
		LL s1 = 0, s2 = 0, s3 = 0, s4 = 0;
		//s4
		memset(C, 0, sizeof(C));
		for(int i = 1; i<= n; i++){
			if(x[i] < y[i]){
				add(y[i], m);
			}
		}
		for(int i = 1; i <= n; i++){
			if(x[i] < y[i]){
				s4 += sum(x[i] - 1);
			}
		}
		//s3 + s4
		memset(C, 0, sizeof(C));
		for(int i = 1; i <= n; i++){
			add(min(x[i], y[i]), m);
		}
		for(int i = 1; i <= n; i++){
            if(x[i] < y[i]){
                s3 += sum(x[i] - 1);
            }
		}

		memset(C, 0, sizeof(C));
		for(int i = 1; i <= n; i++){
			add(max(x[i], y[i]), m);
		}
		for(int i = 1; i <= n; i++){
            if(x[i] < y[i]){
                s3 += n - sum(y[i]);
            }
		}

		s3 -= bit::solve(n);
		s3 -= s4;

		//s1
		memset(C, 0, sizeof(C));
		tot = 0;
		for(int i = 1; i <= n; i++){
			if(x[i] >= y[i]){
                tot++;
				add(y[i], m);
			}
		}
		for(int i = 1; i <= n; i++){
			if(x[i] >= y[i]){
				s1 += tot - sum(x[i] - 1);
			}
		}
		s2 = (LL)n * n - s1 - s3 - s4;
		printf("%I64d %I64d %I64d %I64d\n", s1, s2, s3, s4);
	}

    return 0;
}