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80Days游戏策略算法解析

最新推荐文章于 2018-11-13 21:29:32 发布

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本文探讨了基于儒勒·凡尔纳科幻小说《八十天环游地球》的简化版80Days游戏策略。玩家需管理资金和时间，选择最佳起点城市，确保旅行过程中资金不为负，实现环球旅行目标。文章分析了算法解决方案，包括输入输出格式和示例，为理解游戏策略提供了关键洞察。

#1831 : 80 Days
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
2
3 0
3 4 5
5 4 3
3 100
-3 -4 -5
30 40 50
样例输出
2
-1

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=2e6+5;
long long a[maxn],b[maxn];

int main()
{
    int t;
    scanf("%d",&t);
    long long x;
    int n;
    while(t--)
    {
        queue<long long>q;
        scanf("%d%lld",&n,&x);
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&a[i]);
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&b[i]);
        }
        for(int i=n+1;i<=n*2; i++)
        {
            a[i]=a[i-n];
            b[i]=b[i-n];
        }
        int flag=0;

        for(int i=1; i<=n*2; i++)
        {
            if(x+a[i]-b[i]>=0)
            {
                x+=a[i]-b[i];
                q.push(i);
                if(q.size()>=n)
                {
                    printf("%lld\n",q.front());
                    flag=1;
                    break;
                }
            }
            else
            {
                while(x+a[i]-b[i]<0&&q.size())
                {
                    x-=a[q.front()]-b[q.front()];
                    q.pop();
                }
                if(x+a[i]-b[i]>=0)
                {
                    x+=a[i]-b[i];
                    q.push(i);
                    if(q.size()>=n)
                    {
                        printf("%lld\n",q.front());
                        flag=1;
                        break;
                    }
                }
            }
        }
        if(flag==0)
        {
            printf("-1\n");
        }
    }
}