2018北京网络赛D 80days （尺取）

#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 a_i dollars (a_i can even be negative), and if you want to go to the next city on the circle, you should pay b_i 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, …, a_n  (-10⁹ ≤ a_i ≤ 10⁹), and the third line contains n integers b₁, …, b_n (0 ≤ b_i ≤ 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


比赛的时候都以为是dp，但是自己写的感觉就是暴力。。。

思路：p[i]=a[i]-b[i] 化环为直 尺取法取长度为n的序列，front如果大于n就是-1 （见注释）

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define FO freopen("in.txt", "r", stdin);
#define debug(x) cout << "&&" << x << "&&" << endl;
#define lowbit(x) (x&-x)
#define mem(a,b) memset(a, b, sizeof(a));
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
const int inf = 0x3f3f3f3f;
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//head

const int N=1e6+5;
int p[N<<1],a[N],b[N],_,n;
ll c;
int main() {
    for(scanf("%d",&_);_;_--) {
        scanf("%d%lld",&n,&c);
        rep(i,1,n+1) scanf("%d",&a[i]);
        rep(i,1,n+1) scanf("%d",&b[i]);
        rep(i,1,n+1) p[i]=p[i+n]=a[i]-b[i];//模拟环
        int front=1,rear=1;//都指向1
        while(front<=n&&rear-front+1<=n) {//front<=n n长度
            c+=p[rear];//累加 后移
            rear++;
            while(c<0){//如果c<0 说明front不能到达rear front后移
                c-=p[front];
                front++;
            }
        }
        if(front>n) puts("-1");//1~n都不行
        else printf("%d\n",front);//可以取长度为n的序列 输出队头
    }
}

 

（貌似提交是过不去的。。。）