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=;i<=n;i++) scanf("%d",&a[i]);
         for (int i=;i<=n;i++) scanf("%d",&b[i]);
         for (int i=;i<=n;i++)
         {
             c[i]=a[i]-b[i];
             c[i+n]=c[i];
         }
         int l=,r=;
         long long ans=;
         while (l<=n && r-l+<=n)
         {
             ans+=c[r];
             r++;
             while (ans+m<)
             {
                 ans-=c[l];
                 l++;
             }
         }
         if (l>n) printf("-1\n");else printf("%d\n",l);
     }
     return ;
 }

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+;
 ll a[maxn*];
 ll b[maxn*];
 int n;
 ll sum;
 int main() {
     int caseCnt;
     scanf("%d",&caseCnt);
     while(caseCnt--) {
         scanf("%d%lld",&n,&sum);
         rep(i,,n-) scanf("%lld",&a[i]);
         rep(i,n,*n-) a[i]=a[i-n];
         rep(i,,n-) scanf("%lld",&b[i]);
         rep(i,n,*n-) b[i]=b[i-n];
         int l=;
         while(l<n) {
             if(sum+a[l]>=) {
                 break;
             }
             l++;
         }
         if(l==n) puts("-1");
         else {
             int r=l+;
             ll res=sum+a[l];
             ll temp=res;
             bool ok=false;
             while(r<*n) {
                 if(r-l==n) temp=res-b[r-];
                 else temp=min(res-b[r-],res-b[r-]+a[r]);
                 if(temp<) break;
                 if(r-l==n) {
                     ok=true;
                     break;
                 }
                 res=res-b[r-]+a[r];
                 r++;
             }
             if(ok) {
                 printf("%d\n",l+);
                 continue;
             }
             int ans;
             while(l<n) {
                 while(l<r&&temp<) {
                     temp=temp-a[l]+b[l];
                     res=res-a[l]+b[l];
                     l++;
                 }
                 res=res-b[r-]+a[r];
                 if(l==r) {
                     while(l<n) {
                         if(sum+a[l]>=) {
                             break;
                         }
                         l++;
                     }
                     if(l==n) break;
                     r=l;
                     res=sum+a[l];
                 }
                 r=r+;
                 while(r<*n) {
                     if(r-l==n) temp=res-b[r-];
                     else temp=min(res-b[r-],res-b[r-]+a[r]);
                     if(temp<) break;
                     if(r-l==n) {
                         ans=l+;
                         ok=true;
                         break;
                     }
                     res=res-b[r-]+a[r];
                     r++;
                 }
                 if(ok) break;
             }
             if(!ok) puts("-1");
             else {
                 printf("%d\n",ans);
             }
         }
     }
     return ;
 }

 #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[], b[];
 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 = ; i < n; i++) {
             scanf("%lld", &a[i]);
             a[n + i] = a[i];
         }
         for (i = ; i < n; i++) {
             scanf("%lld", &b[i]);
             b[n + i] = b[i];
         }
         l = ;
         while (l < n&&c + a[l] < )
             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 + ] <  && r +  != (n + l)) {
                 need -= a[r + ];
                 flag = true;
             }
             while (r < (n + l) && res >= need)
             {
                 r++;
                 res = res - need;
                 if(!flag)
                     res = res + a[r];
                 flag = false;
                 if (r == (n + l - )) {
                     need = b[r];
                     flag = false;
                 }
                 else {
                     need = max(b[r], b[r] - a[r+]);
                     if (a[r+] < )
                         flag = true;
                 }
             }
             if (r == (l + n))
             {
                 ok = true;
                 ans = l;
                 break;
             }
             res = res - need;
             r++;
             while (l < r&&res + b[l] - a[l] < ) {
                 res = res + b[l] - a[l];
                 l++;
             }
             if (!flag)
                 res += a[r];
             if (l == r)
             {
                 while (l < n&&c + a[l] < )
                     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 + );
         }
         else
             printf("-1\n");
     }
 }