POJ 1990 MooFest

最新推荐文章于 2021-04-14 16:10:12 发布

原创最新推荐文章于 2021-04-14 16:10:12 发布 · 411 阅读

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文章标签：

#poj #Tree_Array #acm #algorithm

本文介绍了一种利用两棵数状数组解决特定问题的算法实现，通过实例讲解了如何进行数据处理与查询更新操作，适用于解决区间查询与更新问题。

Need two tree array to get answer;

The portal:http://poj.org/problem?id=1990

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>

using namespace std;

const long long  M = 20005;

struct Cows {
	long long  value;
	long long  position;
	bool operator < (const Cows & a1) const {
		return a1.value > value;
	}
}cows[M];

long long  Bit_num[2][M+5];

void Insert(long long  x,long long  value,long long flag) {
	for(long long  i = x ; i <= M ; i += i & -i) {
		Bit_num[flag][i] += value;
	}
}

long long  Query(long long  x,long long flag) {
	long long  ret = 0;
	for(long long  i = x ; i ; i -= i & -i) {
		ret += Bit_num[flag][i];
	}
	return ret;
}

void Deal_with() {
	long long  n;
	while(~scanf("%lld",&n)){
		memset(Bit_num,0,sizeof(Bit_num));
		for(long long  i=1 ; i <= n ; i++) {
			scanf("%lld %lld",&cows[i].value,&cows[i].position);
		}
		sort(cows+1,cows+n+1);
		long long  ans = 0;
		for(long long  i=1 ; i <= n ; i++) {
			long long  a = Query(cows[i].position,0); 
			long long  b = Query(cows[i].position,1);
			long long  c = Query(M,1);
			ans += cows[i].value * (cows[i].position * a - b +
			c - b - (i - a - 1) * cows[i].position) ;
			Insert(cows[i].position,cows[i].position,1);
			Insert(cows[i].position,1,0);
		}
		printf("%lld\n",ans);
	}
}

int main(void) {
	//freopen("a.in","r",stdin);
	Deal_with();
	return 0;
}