[Nowcoder 2018ACM多校第二场K] carpet

题目大意：给你一个n*m的矩阵A，由’a’-‘z’组成，除此以外，每个位置还有个权值cost[i][j]。求一个p*q子矩阵B，使得B复制成无穷份构成一个无穷大的矩阵后， A是其子矩阵，并输出最小的费用。定义矩阵B的费用是其每个元素的最大权值 * (p + 1) * (q + 1)。 $(n * m \leq 10^6, cost[i][j] \leq 10^9)$

题目思路：首先可以解决第一个问题，找出矩阵A的循环节，这个可以对行和列分别来处理，将矩阵分别按行哈希和按列哈希再根据kmp找到各自的最小循环节，得到p和q。可以知道矩阵A的所有p * q的子矩阵都是满足要求的，然后找费用最小的子矩阵，也就是维护每个p * q子矩阵的最大元素，这个可以通过两遍单调队列来维护。

PS：这里的循环节是可以允许母串还剩一段的，即最小循环节长度就是n-next[n]，不需要一般情况下n%(n-next[n])==0的条件。

Code:

#include <map>
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define ll long long
#define id(i, j) (((i) - 1) * m + j)

const int N = (int)1e6 + 10;

int n, m;
char str[N]; int w[N];
int f[N], head, tail, que[N][2];

unsigned int t[N];

int gcd(int a, int b){
    if (b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(int a, int b){
    return 1ll * a * b / gcd(a, b);
}
int nxt[N];

int main(){

    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i ++){
        scanf("%s", str + (i - 1) * m + 1);
    }

    for (int i = 1; i <= n * m; i ++)
        scanf("%d", w + i);

    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++)
            t[i] = t[i] * 177 + str[id(i, j)];

    int p, q;
    {
        int j = 0;
        for (int i = 2; i <= n; i ++){
            while (t[j + 1] != t[i] && j > 0) j = nxt[j];
            if (t[j + 1] == t[i]) j ++;

            nxt[i] = j;
        }

        p = n - nxt[n];
    }

    for (int j = 1; j <= m; j ++){
        t[j] = 0;
        for (int i = 1; i <= n; i ++)
            t[j] = t[j] * 177 + str[id(i, j)];
    }

    {
        int j = 0;
        for (int i = 2; i <= m; i ++){
            while (t[j + 1] != t[i] && j > 0) j = nxt[j];
            if (t[j + 1] == t[i]) j ++;

            nxt[i] = j;
        }   

        q = m - nxt[m];
    }
    printf("%d %d\n", p, q);

    // p*q subrectangle
    for (int j = 1; j <= m; j ++){
        tail = head = 1;
        que[1][0] = w[id(1, j)], que[1][1] = 1;

        for (int i = 2; i <= p; i ++){
            while (head <= tail && que[tail][0] <= w[id(i, j)]) tail --;
            ++ tail;
            que[tail][0] = w[id(i, j)], que[tail][1] = i;
        }


        for (int i = 1; i <= n - p + 1; i ++){
            if (que[head][1] < i) head ++;
            f[id(i, j)] = que[head][0];

            if (i + p <= n){
                while (head <= tail && que[tail][0] <= w[id(i + p, j)]) tail --;
                ++ tail;
                que[tail][0] = w[id(i + p, j)], que[tail][1] = i + p;       
            }
        }
    }

    int ans = 1 << 30;
    for (int i = 1; i <= n - p + 1; i ++){
        head = tail = 1;
        que[1][0] = f[id(i, 1)], que[1][1] = 1;

        for (int j = 2; j <= q; j ++){
            while (head <= tail && que[tail][0] <= f[id(i, j)]) tail --;
            ++ tail;
            que[tail][0] = f[id(i, j)], que[tail][1] = j;
        }

        for (int j = 1; j <= m - q + 1; j ++){
            if (que[head][1] < j) head ++;

            ans = min(ans, que[head][0]);

            if (j + q <= m){
                while (head <= tail && que[tail][0] <= f[id(i, j + q)]) tail --;
                ++ tail;
                que[tail][0] = f[id(i, j + q)], que[tail][1] = j + q;
            }
        }

    }


    printf("%lld\n", 1ll * ans * (p + 1) * (q + 1));

    return 0;
}

[Nowcoder 2018ACM多校第二场K] carpet

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