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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#567271#9313. Make MaxZhumingTL 4ms16008kbC++174.3kb2024-09-16 10:43:112024-09-16 10:43:12

Judging History

你现在查看的是最新测评结果

  • [2024-09-18 15:56:24]
  • hack成功,自动添加数据
  • (/hack/836)
  • [2024-09-16 10:43:12]
  • 评测
  • 测评结果:TL
  • 用时:4ms
  • 内存:16008kb
  • [2024-09-16 10:43:11]
  • 提交

answer

#include <iostream>
#include <cstring>
using namespace std;
#define ll long long 
const int N = 3e5+10;
#include <cstring>
ll a[N];
ll tree[N<<2];
ll tag[N<<2];
ll Same[N<<2];
ll ls(ll p){return p<<1;}
ll rs(ll p){return p<<1|1;}
inline void build(ll p,ll pl,ll pr){
    tag[p] = 0;
    Same[p] = 0;
    if(pl==pr) {tree[p]=a[pl];Same[p]=1;return;}
    ll mid = (pl+pr) >> 1;
    build(ls(p),pl,mid);
    build(rs(p),mid+1,pr);
    tree[p] = min(tree[ls(p)],tree[rs(p)]);
    if(tree[ls(p)]==tree[rs(p)]&&Same[ls(p)]==1&&Same[rs(p)]==1){
        Same[p] = 1;
    }
}

inline void update(ll L,ll R,ll p,ll pl,ll pr,ll d){
    if(L<=pl&&pr<=R){
        tag[p] = d;
        tree[p] = d;
        return;
    }
    if(tag[p]){
        ll mid = (pl+pr) >> 1;
        tag[ls(p)] = tag[p];
        tree[ls(p)] = tag[p];
        tag[rs(p)] = tag[p];
        tree[rs(p)] = tag[p];
        tag[p] = 0;
    }
    ll mid = (pl+pr) >> 1;
    if(L<=mid) update(L,R,ls(p),pl,mid,d);
    if(R>mid) update(L,R,rs(p),mid+1,pr,d);
    tree[p] = min(tree[ls(p)],tree[rs(p)]);
    if(tree[ls(p)]==tree[rs(p)]&&Same[ls(p)]==1&&Same[rs(p)]==1){
        Same[p] = 1;
    }
}
inline ll query(ll p,ll pl,ll pr){
    //先找到两个字段谁是最小
    if(pl==pr) return pl;//返回这个位置
    if(tag[p]){
        ll mid = (pl+pr) >> 1;
        tag[ls(p)] = tag[p];
        tree[ls(p)] = tag[p];
        tag[rs(p)] = tag[p];
        tree[rs(p)] = tag[p];
        tag[p] = 0;
    }
    ll mid = (pl+pr) >> 1;
    if(tree[ls(p)]<=tree[rs(p)]){
        return query(ls(p),pl,mid);
    }else{
        return query(rs(p),mid+1,pr);
    }
}
inline ll query2(ll p,ll pl,ll pr){
    ll mid = (pl+pr) >> 1;
    if(pl==pr){
        return pl;
    }
    if(tag[p]){
        ll mid = (pl+pr) >> 1;
        tag[ls(p)] = tag[p];
        tree[ls(p)] = tag[p];
        tag[rs(p)] = tag[p];
        tree[rs(p)] = tag[p];
        tag[p] = 0;
    }
    if(tree[ls(p)]<tree[rs(p)]){//那就从左边开始找
        if(Same[ls(p)]==1){
            return mid;
        }
        return query2(ls(p),pl,mid);
    }else if(tree[ls(p)]>tree[rs(p)]){
        if(Same[rs(p)]==1){
            return pr;
        }
        return query2(rs(p),mid+1,pr);
    }else if(tree[ls(p)]==tree[rs(p)]){
        //那就从两个中都找一遍
        //如果左边都是一样的,就看右边,如果左边不一样就看左边
        if(Same[p]==1){
            return pr;
        }
        if(Same[ls(p)]==1&&Same[rs(p)]==0){

        }
        if(Same[ls(p)]==1){//如果都相同,就看右边,如果不相同再看左边
            return query2(rs(p),mid+1,pr);
        }else{
            return query2(ls(p),pl,mid);
        }
    }
}
inline ll qMin(ll p,ll pl,ll pr,ll pos){
    while(pl!=pr){
        if(tag[p]){
            ll mid = (pl+pr) >> 1;
            tag[ls(p)] = tag[p];
            tree[ls(p)] = tag[p];
            tag[rs(p)] = tag[p];
            tree[rs(p)] = tag[p];
            tag[p] = 0;
        }
        ll mid = (pl+pr) >> 1;
        if(pos>=pl&&pos<=mid){
            p = ls(p);
            pr =  mid;
        }else{
            p = rs(p);
            pl = mid+1;
        }
    }
    return tree[p];
}
int main()
{
    int t;scanf("%d",&t);
    while(t--){
        memset(tag,0,sizeof(tag));
        int n;scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
        build(1,1,n);
        ll Ans = 0;
        while(1){
            //找到最小值,和它的左右
            ll left = query(1,1,n);
            ll right = query2(1,1,n);
            //找到左右的旁边两个中的最小值
            if(left==1&&right==n) break;
            if(left==1){
                //取pl == right+1的数据
                ll Min = qMin(1,1,n,right+1);
                update(left,right,1,1,n,Min);
                Ans += (right - left + 1);
                continue;
            }
            if(right==n){
                ll Min = qMin(1,1,n,left-1);
                update(left,right,1,1,n,Min);
                Ans += (right - left + 1);
                continue;
            }
            ll Min = min(qMin(1,1,n,left-1),qMin(1,1,n,right+1));
            update(left,right,1,1,n,Min);
            Ans += (right - left + 1);
        }
        cout << Ans << endl;
    }
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 4ms
memory: 16008kb

input:

4
2
1 2
2
2 2
7
1 1 1 2 2 2 2
3
1 2 3

output:

1
0
3
3

result:

ok 4 number(s): "1 0 3 3"

Test #2:

score: -100
Time Limit Exceeded

input:

2
198018
875421126 585870339 471894633 383529988 625397685 944061047 704695631 105113224 459022561 760848605 980735314 847376362 980571959 329939331 644635272 326439858 752879510 837384394 175179068 182094523 397239381 1199016 185143405 279638454 252374970 822030887 860312140 137248166 993229443 164...

output:


result: