CodeForces 253D Table with Letters - 2 减小复杂度的技巧+交替枚举

E - Table with Letters - 2

Time Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatus

Description

input

input.txt

output

output.txt

Vasya has recently started to learn English. Now he needs to remember how to write English letters. He isn't sure about some of them, so he decided to train a little.

He found a sheet of squared paper and began writing arbitrary English letters there. In the end Vasya wroten lines containing m characters each. Thus, he got a rectangular n × m table, each cell of the table contained some English letter. Let's number the table rows from top to bottom with integers from 1 ton, and columns — from left to right with integers from 1 tom.

After that Vasya looked at the resulting rectangular table and wondered, how many subtables are there, that matches both following conditions:

• the subtable contains at most k cells with "a" letter;
• all letters, located in all four corner cells of the subtable, are equal.

Formally, a subtable's definition is as follows. It is defined by four integersx₁, y₁, x₂, y₂ such that1 ≤ x₁ < x₂ ≤ n,1 ≤ y₁ < y₂ ≤ m. Then the subtable contains all such cells(x, y) (x is the row number,y is the column number), for which the following inequality holdsx₁ ≤ x ≤ x₂, y₁ ≤ y ≤ y₂. The corner cells of the table are cells (x₁, y₁),(x₁, y₂),(x₂, y₁),(x₂, y₂).

Vasya is already too tired after he's been writing letters to a piece of paper. That's why he asks you to count the value he is interested in.

Input

The first line contains three integers n, m, k(2 ≤ n, m ≤ 400; 0 ≤ k ≤ n·m).

Next n lines contain m characters each — the given table. Each character of the table is a lowercase English letter.

Output

Print a single integer — the number of required subtables.

Sample Input

Input

3 4 4aabbbaabbaab

Output

2

Input

4 5 1ababaccacaccacbcbabc

Output

1

Hint

There are two suitable subtables in the first sample: the first one's upper left corner is cell(2, 2) and lower right corner is cell (3, 3), the second one's upper left corner is cell (2, 1) and lower right corner is cell (3, 4).

解题思路：

这题刚开始做的时候想到的只是复杂度为n^4的算法，codeforce上肯定超时

那么这题如何解决？要么是一个高效的数据结构来降低复杂度，要么就是技巧（枚举的技巧，标记的技巧等）

1，首先对于K的判断，求一个矩阵范围内的a的数量那就是递推啊，这个算做是基础吧，n^2解决

2，枚举的策略是枚举列。每次枚举两列，然后再这两行中，交替进行列的枚举，同时使用标记，找一个样例手推一遍代码就知道为什么了。

3，总的复杂度n^3还是很好的解决方案的，做过一遍这样的题，下一次就不慌了。

#include<cstdio>#include<cstring>#include<iostream>using namespace std;const int maxn = 500 ;int n,m,K;char mapp[maxn][maxn] ;int num[maxn][maxn] ;int main(){    freopen("input.txt", "r", stdin);    freopen("output.txt", "w", stdout);    while(~scanf("%d%d%d",&n,&m,&K)){        for(int i=1;i<=n;i++){            scanf("%s",mapp[i]+1);        }        memset(num,0,sizeof(num));        for(int i=1;i<=n;i++){            for(int j=1;j<=m;j++){                num[i][j] = num[i-1][j]+num[i][j-1]-num[i-1][j-1] ;                if(mapp[i][j]=='a')num[i][j]++;            }        }//        for(int i=1;i<=n;i++){//            for(int j=1;j<=m;j++){//                printf("%d ",num[i][j]);//            }printf("\n");//        }        long long ans = 0;        int flag[500] ;        for(int i=1;i<=n;i++){            for(int j = i+1;j<=n;j++){                int p = 1 ;                memset(flag,0,sizeof(flag));                for(int k = 1;k<=m;k++){                    if(mapp[i][k]!=mapp[j][k])continue ;                    flag[mapp[i][k]]--;                    while(p<=m&&num[j][p]-num[i-1][p]-num[j][k-1]+num[i-1][k-1]<=K){                        if(mapp[i][p]==mapp[j][p])flag[mapp[i][p]]++; ;                        p++;                    }                    if(flag[mapp[i][k]]>0)ans+=flag[mapp[i][k]];                }            }        }        printf("%I64d\n",ans);    }    return 0;}