uva 11054 Gerovia的酒交易（贪心+树状数组）

标签：贪心   树状数组   

直线上有n个等距的村庄，每个村庄要么买酒，要么卖酒。把k个单位的酒从一个村庄运到相邻村庄需要k个单位的劳动力。问最少需要多少劳动力才能满足所有村庄的需求

思路贪心

#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<cstdlib>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<queue>  
#include<stack> 
#include<string>
#include<map> 
#include<set>
#define eps 1e-6 
#define LL long long  
using namespace std;  

const int maxn = 100000 + 100;
const int INF = 0x3f3f3f3f;
//freopen("input.txt", "r", stdin);
int C[maxn], n;

int lowbit(int x) { return x&(-x);}
int sum(int x) {
	if(x == 0) return 0;
	int ret = 0;
	while(x > 0) {
		ret += C[x]; x -= lowbit(x);
	}
	return ret;
}
void add(int x, int d) {
	while(x <= n) {
		C[x] += d; x += lowbit(x);
	}
}

LL solve(int left, int right) {
	if(left == right) return (LL)0;
	if(left+1 == right) return (LL)abs(sum(right)-sum(left));
	int mid = (left + right) >> 1;
	int tmp = sum(mid) - sum(left-1);
	add(mid, -tmp); add(mid+1, tmp);
	return solve(left, mid) + solve(mid+1, right) + (LL)abs(tmp); 
}

void init() {
	memset(C, 0, sizeof(C));
	int tmp;
	for(int i = 1; i <= n; i++) {
		cin >> tmp;
		add(i, tmp); 
	}
}

int main() {
	//freopen("input.txt", "r", stdin);
	while(scanf("%d", &n) == 1 && n) {
		init();
		cout << solve(1, n) << endl;
	}
	return 0;
}

uva 11054 Gerovia的酒交易（贪心+树状数组）

标签：贪心   树状数组   

原文地址：http://blog.csdn.net/u014664226/article/details/46503115