ACM-ICPC 2018 北京赛区网络预赛 80 Days

最新推荐文章于 2018-12-29 13:35:01 发布

原创最新推荐文章于 2018-12-29 13:35:01 发布 · 355 阅读

0 ·

CC 4.0 BY-SA版权

枚举法尺取法专栏收录该内容

2 篇文章

订阅专栏

本文探讨基于儒勒·凡尔纳科幻小说《八十天环游世界》的简化游戏80Days，玩家需管理资金与时间，选择最优起始城市完成环球旅行。文章通过实例分析，介绍了使用滑动窗口法解决此问题的算法思路。

时间限制:1000ms

单点时限:1000ms

内存限制:256MB

描述

80 Days is an interesting game based on Jules Verne's science fiction "Around the World in Eighty Days". In this game, you have to manage the limited money and time.

Now we simplified the game as below:

There are n cities on a circle around the world which are numbered from 1 to n by their order on the circle. When you reach the city i at the first time, you will get ai dollars (ai can even be negative), and if you want to go to the next city on the circle, you should pay bi dollars. At the beginning you have c dollars.

The goal of this game is to choose a city as start point, then go along the circle and visit all the city once, and finally return to the start point. During the trip, the money you have must be no less than zero.

Here comes a question: to complete the trip, which city will you choose to be the start city?

If there are multiple answers, please output the one with the smallest number.

输入

The first line of the input is an integer T (T ≤ 100), the number of test cases.

For each test case, the first line contains two integers n and c (1 ≤ n ≤ 106, 0 ≤ c ≤ 109). The second line contains n integers a1, …, an (-109 ≤ ai ≤ 109), and the third line contains n integers b1, …, bn (0 ≤ bi ≤ 109).

It's guaranteed that the sum of n of all test cases is less than 106

输出

For each test case, output the start city you should choose.

提示

For test case 1, both city 2 and 3 could be chosen as start point, 2 has smaller number. But if you start at city 1, you can't go anywhere.

For test case 2, start from which city seems doesn't matter, you just don't have enough money to complete a trip.

样例输入

样例输出

2
-1

题意：n个成环的城市，到达第i个城市获得ai ，离开丢掉bi ，初始有c，全程拥有值不能为负数，问是否可行，可行就输出最小的出发点。

先用倍增法扩大数组长度，考试时，想到用尺取法，一心思的想维护这段区间的最小值，当区间的最小值处，满足，但是，接着end往后加，加到这不满足了，说明当前位置比前面维护的最小值更小，所以这时候就让减去a[star],star++,就行了；当时就是一心的想维护前面的最小值，导致最终也没写出来，说明思考的不透彻；

代码：

#include<bits/stdc++.h>
using namespace std;
#define Max 1000005
#define ll long long 
ll a[2*Max],b[Max],c;
int n;
int main()
{
	int t;
 	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%lld",&n,&c);
		for(int i= 1;i<=n;i++)
			scanf("%lld",&a[i]);
		for(int i = 1;i<=n;i++)
		{
			scanf("%lld",&b[i]);
			a[i] = a[i] - b[i];
			a[i+n] = a[i];
		}
		ll sum = c;
		for(int i = 1;i <= n;i ++)
			sum+= a[i];
		if(sum<0)
		{
			printf("-1\n");
			continue;
		}
		sum = 0;
		int star = 1,end = 1,flag = 0;
		while(end<=2*n)
		{
			if(sum+c<0)
			{
				sum -= a[star];
				star++;
				if(star>end)
					end = star;
			}
			else
			{
				sum += a[end];
				if(end-star+1 == n&& sum+c>=0)
				{
					flag = star;
					break;
				}
				end++;
			}		
		}
		if(flag) printf("%d\n",flag);
		else printf("-1\n");
	}
	return 0;
}