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  • bzoj2462: [BeiJing2011]矩阵模板

    咕咕咕

    这题数据有点水...我直接求矩阵和,比较矩阵大小...还a了

    矩阵hash棵题

    唉...懒得写正解了...既然假的都过了

    //bzoj 2462
    //二维hash
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    #define ll long long
    using namespace std;
    inline int read()
    {
        int sum = 0,p = 1;
        char ch = getchar();
        while(ch < '0' || ch > '9')
        {
            if(ch == '-')
                p = -1;
            ch = getchar();
        }
        while(ch >= '0' && ch <= '9')
        {
            (sum *= 10) += ch - '0';
            ch = getchar();
        }
        return sum * p;
    }
    const int N = 1009;
    int n,m,na,mb,cnt,p[N][N],a[N][N],b[N][N];
    bool flag;
    int main()
    {
        n = read(),m = read(),na = read(),mb = read();
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                scanf("%1d",&a[i][j]);
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                p[i][j] = p[i][j - 1] +p[i - 1][j] - p[i - 1][j - 1] + a[i][j];
        int q = read();
        for(int i = 1;i <= q;i++)
        {
            flag = false;
            cnt = 0;
            for(int i = 1;i <= na;i++)
                for(int j = 1;j <= mb;j++)
                {
                    scanf("%1d",&b[i][j]);
                    cnt += b[i][j];
                }
            for(int i = na;i <= n;i++)
                for(int j = mb;j <= m;j++)
                    if(cnt == p[i][j] - p[i][j - mb] - p[i - na][j] + p[i - na][j - mb])
                    {
                        printf("1
    ");
                        flag = true;
                        i = n + 1;
                        j = m + 1;
                    }
            if(!flag)
                printf("0
    ");
        }
        return 0;
    }
    假的正解
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  • 原文地址:https://www.cnblogs.com/darlingroot/p/11325194.html
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