2018亚洲区预选赛北京赛站网络赛 D.80 Days 尺取

题面

题意:你带着K元要去n个城市,这n个城市是环形的,你可以选择任意一个起点,然后顺时针走,对于每个城市,到达时可以获得a元,但是从这里离开又需要花费b元,问你能否找到一个起点(输出花钱最少的那个),使得你能够走完一圈,不能输出-1

题解:首先对于环形问题,先把数组复制一次,现在从每个起点开始,满足条件的点就肯定可以进队,其实我们要求的一个i满足,从i到n+i的所有前缀和最小值一定要大于K,因为对于1个i只会进队一次出队一次,所以尺取一下

(我们一路判断的就是,带着K元能不能加上这段区间>=0,不能的话起点就要顺延,同时终点也顺延,每次只改变头首2个元素)

#include<bits/stdc++.h>
#define N 2000010 
using namespace std;
int T,ans,n,m,cnt,b[N],a[N],c[N];
int main()
{
    scanf("%d",&T);
    while (T--) 
    {
        scanf("%d%d",&n,&m);
        for (int i=1;i<=n;i++) scanf("%d",&a[i]);
        for (int i=1;i<=n;i++) scanf("%d",&b[i]);
        for (int i=1;i<=n;i++)
        {
            c[i]=a[i]-b[i];
            c[i+n]=c[i];
        }
        int l=1,r=1;
        long long ans=0;
        while (l<=n && r-l+1<=n)
        {    
            ans+=c[r];
            r++;
            while (ans+m<0) 
            {
                ans-=c[l];
                l++;
            }
        }
        if (l>n) printf("-1\n");else printf("%d\n",l);
    }
    return 0;
}

 

比赛时,队友为了实现类似想法,却苦于没听过尺取(我也没听过),和不是很能确定a[i]和b[i]是否能统一的问题,大力码的2种。

#include<bits/stdc++.h>
using namespace std;
#define mem(a,i) memset(a,i,sizeof(a))
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define lowbit(x) (x&-x)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
typedef long long ll;
typedef pair<int,int> pii;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+5;
ll a[maxn*2];
ll b[maxn*2];
int n;
ll sum;
int main() {
    int caseCnt;
    scanf("%d",&caseCnt);
    while(caseCnt--) {
        scanf("%d%lld",&n,&sum);
        rep(i,0,n-1) scanf("%lld",&a[i]);
        rep(i,n,2*n-1) a[i]=a[i-n];
        rep(i,0,n-1) scanf("%lld",&b[i]);
        rep(i,n,2*n-1) b[i]=b[i-n];
        int l=0;
        while(l<n) {
            if(sum+a[l]>=0) {
                break;
            }
            l++;
        }
        if(l==n) puts("-1");
        else {
            int r=l+1;
            ll res=sum+a[l];
            ll temp=res;
            bool ok=false;
            while(r<2*n) {
                if(r-l==n) temp=res-b[r-1];
                else temp=min(res-b[r-1],res-b[r-1]+a[r]);
                if(temp<0) break;
                if(r-l==n) {
                    ok=true;
                    break;
                }
                res=res-b[r-1]+a[r];
                r++;
            }
            if(ok) {
                printf("%d\n",l+1);
                continue;
            }
            int ans;
            while(l<n) {
                while(l<r&&temp<0) {
                    temp=temp-a[l]+b[l];
                    res=res-a[l]+b[l];
                    l++;
                }
                res=res-b[r-1]+a[r];
                if(l==r) {
                    while(l<n) {
                        if(sum+a[l]>=0) {
                            break;
                        }
                        l++;
                    }
                    if(l==n) break;
                    r=l;
                    res=sum+a[l];
                }
                r=r+1;
                while(r<2*n) {
                    if(r-l==n) temp=res-b[r-1];
                    else temp=min(res-b[r-1],res-b[r-1]+a[r]);
                    if(temp<0) break;
                    if(r-l==n) {
                        ans=l+1;
                        ok=true;
                        break;
                    }
                    res=res-b[r-1]+a[r];
                    r++;
                }
                if(ok) break;
            }
            if(!ok) puts("-1");
            else {
                printf("%d\n",ans);
            }
        }
    }
    return 0;
}

#include <cstring>
#include <cstdio>
#include<assert.h>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include<iostream>
#include<queue>
#include<functional>
#include <vector>
#include<set>
#include<string>
#include<map>
#include<unordered_set>
using namespace std;
long long a[2000050], b[2000050];
int main() {
    long long i, j, k, n, m, l, r, res, t, c, ans;
    scanf("%lld", &t);
    while (t--)
    {
        scanf("%lld%lld", &n, &c);
        for (i = 0; i < n; i++) {
            scanf("%lld", &a[i]);
            a[n + i] = a[i];
        }
        for (i = 0; i < n; i++) {
            scanf("%lld", &b[i]);
            b[n + i] = b[i];
        }
        l = 0;
        while (l < n&&c + a[l] < 0)
            l++;
        if (l == n)
        {
            printf("-1\n");
            continue;
        }
        r = l; res = c + a[l];
        bool ok = false;
        while (l < n)
        {
            bool flag = false;
            long long need = b[r];
            if (a[r + 1] < 0 && r + 1 != (n + l)) {
                need -= a[r + 1];
                flag = true;
            }
            while (r < (n + l) && res >= need)
            {
                r++;
                res = res - need;
                if(!flag)
                    res = res + a[r];
                flag = false;
                if (r == (n + l - 1)) {
                    need = b[r];
                    flag = false;
                }
                else {
                    need = max(b[r], b[r] - a[r+1]);
                    if (a[r+1] < 0)
                        flag = true;
                }
            }
            if (r == (l + n))
            {
                ok = true;
                ans = l;
                break;
            }
            res = res - need;
            r++;
            while (l < r&&res + b[l] - a[l] < 0) {
                res = res + b[l] - a[l];
                l++;
            }
            if (!flag)
                res += a[r];
            if (l == r)
            {
                while (l < n&&c + a[l] < 0)
                    l++;
                if (l == n)
                {
                    break;
                }
                else
                {
                    r = l;
                    res = c + a[l];
                }
            }
            else
            {
                res = res + b[l] - a[l];
                l++;
            }
        }
        if (ok)
        {
            printf("%lld\n", ans + 1);
        }
        else
            printf("-1\n");
    }
}