G. Mischievous Shooter

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-062.C. Removal of Unattractive Pairs2023-12-063.D. Jumping Through Segments2023-12-084.E. Good Triples2023-12-085.F. Shift and Reverse2023-12-086.D. Yet Another Monster Fight2023-12-117.A. Constructive Problems2023-12-188.C. Game with Multiset2023-12-199.A. Rating Increase2023-12-1910.B. Swap and Delete2023-12-1911.A. Problemsolving Log2023-12-2012.B. Preparing for the Contest2023-12-2013.C. Quests2023-12-2014.D. Three Activities2023-12-2015.E2. Game with Marbles (Hard Version)2023-12-2016.cf刷题有感2023-12-2017.A. Anonymous Informant2023-12-2118.A. Forked!2023-12-2319.B. Make Almost Equal With Mod2023-12-2720.C. Heavy Intervals2023-12-2721.D. Split Plus K2023-12-2822.A. 20232023-12-3123.B. Two Divisors2023-12-3124.C. Training Before the Olympiad2023-12-3125.D. Mathematical Problem2024-01-0226.F. Greetings2024-01-0227.C. Partitioning the Array2024-01-1528.G. Bicycles2024-01-1629.E. Eat the Chip2024-01-1730.G. Lights2024-01-1731.D. Array Repetition2024-01-1732.D. Berserk Monsters2024-01-2233.E. Increasing Subsequences2024-01-2334.D. Very Different Array2024-01-24

35.G. Mischievous Shooter2024-01-25

36.B. Plus-Minus Split2024-01-2737.B. A Balanced Problemset2024-01-2838.C. Did We Get Everything Covered2024-01-2839.D. Find the Different Ones!2024-02-0840.C. Grouping Increases2024-02-1041.D. Good Trip2024-02-1242.C. Physical Education Lesson2024-02-1243.E. Final Countdown2024-02-1844.D. Divisible Pairs2024-02-1945.G. Vlad and Trouble at MIT2024-02-2046.A. Brick Wall2024-02-2847.B. Minimize Inversions2024-02-2848.C. XOR-distance2024-02-2849.A. Moving Chips2024-02-2850.B. Monsters Attack!2024-02-2851.C. Find B2024-02-2852.D. Slimes2024-02-2853.C. Turtle Fingers: Count the Values of k2024-02-2854.D. Turtle Tenacity: Continual Mods2024-02-2855.D. Vlad and Division2024-02-2856.C. LR-remainders2024-02-2857.D. Lonely Mountain Dungeons2024-03-12

收起

原题链接

思路简述

二维差分+矩阵旋转

思路详述

1.二维差分，对于每一个标签而言，有对一维的影响和二维的传递之分
2.为什么要差分？对于每一个目标而言，它对以其为左上角顶点，k为边长的三角形内的点都有一个贡献，这种范围内的累加就考虑用前缀和（这里是二维差分）
3.为什么要矩阵旋转？由于在某个点喷的方向不同，得到的值也不同。所以每个目标的贡献只能对一个方向
4.好烦啊，怎么用文字表述呢

$Code$

#include<bits/stdc++.h>
using namespace std;
int n,m,k;
string s[100005];
int ss1()
{
    vector<vector<int> > a(n+5,vector<int> (m+5));//二维数组n*m
    vector<vector<int> > b(n+5,vector<int> (m+5));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(s[i][j]=='#')
            {
                a[i][j]++;
                if(i+k<n)
                {
                    a[i+k+1][j]--;//传递截至点
                    b[i+k+1][j]++;
                }
                if(j+k<m) b[i][j+k+1]--;//b代表斜线传递
                else if(i+j+k-m+1<=n) b[i+j+k-m+1][m]--;//三角形的范围可能超过了矩阵范围
            }
        }
    }

    int ans=0;
    for(int i=1;i<=n;i++)
    {
        int sum=0;
        for(int j=1;j<=m;j++)
        {
            sum+=a[i][j]+b[i][j];//赋能
            ans=max(ans,sum);
            a[i+1][j]+=a[i][j];//传递
            b[i+1][j-1]+=b[i][j];
        }
    }
    return ans;
}

void rotate90Degrees()
{
    vector<string> temp(m+1);//二维字符串，string是一维，vector是一维，m+1代表vector的大小，即含有几个string
    for (int j = 1; j <= m; ++j) temp[j].resize(n+1, ' ');//把temp中的每一个元素（string）变成长度为n+1的空格

    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            temp[j][n-i+1] = s[i][j];//旋转，注意这里不能temp[i][j]=s[j][n-i+1],因为s会越界

    for (int i = 1; i <= m; i++) s[i] = temp[i];//字符串的复制可以直接等于
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {

        cin>>n>>m>>k;
        for(int i=1;i<=n;i++)
        {
            cin>>s[i];
            s[i]=" "+s[i];//使下标从1开始
        }

        int ans=0;
        for(int i=1;i<=4;i++)
        {
            ans=max(ans,ss1());
            rotate90Degrees();
            swap(n,m);//旋转后的矩阵nm发生变换
        }
        cout<<ans<<endl;
    }
    return 0;
}