P4147 玉蟾宫

合集 - 思维题(99)

1.大雪球2024-01-162.E. Final Countdown2024-02-183.D. Divisible Pairs2024-02-194.F. Chat Screenshots2024-02-195.G. One-Dimensional Puzzle2024-02-206.P8775 [蓝桥杯 2022 省 A] 青蛙过河2024-02-217.Watering an Array2024-02-238.Dora and Search2024-02-249.P1137 旅行计划2024-02-2410.P3067 [USACO12OPEN] Balanced Cow Subsets G2024-02-2611.P4799 [CEOI2015 Day2] 世界冰球锦标赛2024-02-2612.A. Brick Wall2024-02-2813.B. Minimize Inversions2024-02-2814.C. XOR-distance2024-02-2815.A. Moving Chips2024-02-2816.B. Monsters Attack!2024-02-2817.C. Find B2024-02-2818.D. Slimes2024-02-2819.C. Turtle Fingers: Count the Values of k2024-02-2820.D. Turtle Tenacity: Continual Mods2024-02-2821.D. Vlad and Division2024-02-2822.C. LR-remainders2024-02-2823.P5322 [BJOI2019] 排兵布阵2024-03-0324.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-0325.P1156 垃圾陷阱2024-03-0326.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-0427.P3047 [USACO12FEB] Nearby Cows G2024-03-0428.P5020 [NOIP2018 提高组] 货币系统2024-03-0429.P3957 [NOIP2017 普及组] 跳房子2024-03-0630.P1550 [USACO08OCT] Watering Hole G2024-03-0631.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0732.P1379 八数码难题2024-03-0733.P6121 [USACO16OPEN] Closing the Farm G2024-03-0834.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0835.P9632 [ICPC2020 Nanjing R] K Co-prime Permutation2024-03-0836.P9825 [ICPC2020 Shanghai R] Fibonacci2024-03-0837.P2034 选择数字2024-03-0938.P7072 [CSP-J2020] 直播获奖2024-03-0939.P1714 切蛋糕2024-03-09

40.P4147 玉蟾宫2024-03-10

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收起

原题链接

题解

太巧妙了

把每个点上方的连续f长度记录下来，然后求每行的柱状图构成的矩形的最大面积

code

#include<bits/stdc++.h>
using namespace std;
int f[1005][1005]={0};
int n,m;

struct node
{
    int h,cnt;
};

int solve(int row)//每一行列上的高度，如果全部使用，不是作为左端点就是作为右端点
{
    int ans=0;
    stack<node> q;//相当于求左边最远小于自己的值，右边最远小于自己的值，单调栈解法
    for(int j=1;j<=m;j++)
    {
        int cnt=0;
        while(q.size()&&q.top().h>=f[row][j])
        {
            cnt+=q.top().cnt;//cnt代表这个点收纳了多少点
            ans=max(ans,q.top().h*cnt);//作为左端点更新，即后边遇到了比自己更小的值，然后弹出
            q.pop();//如果弹出，代表遇到了作为左端点的最远距离的点
        }

        ans=max(ans,f[row][j]*(cnt+1));//作为右端点更新，找到左边比自己小的点
        q.push({f[row][j],cnt+1});//cnt代表该元素左边有多少不小于自己的(包括自己)
    }

    int cnt=0;
    while(q.size())//有些元素没有作为左端点更新过
    {
        ans=max(ans,(cnt+1)*q.top().h);
        cnt+=q.top().cnt;
        q.pop();
    }
    return ans;
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            char op;
            cin>>op;
            if(op=='F') f[i][j]=f[i-1][j]+1;//记录以该点为下端点，最长连续f的长度
        }
    }

    int ans=0;
    for(int i=1;i<=n;i++)
    {
        ans=max(ans,solve(i));//变成求每行的最大矩形
    }

    cout<<3*ans;
    return 0;
}