Sicily 1914. Kingdoms

1914. Kingdoms

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

 The former king of Gridland - CR - has just passed away, leaving his K sons that all desire the throne. Then the war of succession breaks out and the K -King era begins.

Gridland is a rectangle of N * M , with each grid unit as a city. Each city supports one and only one of the K kings as its leader. Cities supporting the same king will always endeavor to establish a kingdom through fighting the rival, and they will not stop until they establish a kingdom satisfying the following rules:

1. The kingdom is a rectangle on the map.
2. Every city on the Gridland supporting that king should be inside the kingdom.

Sadly, through the process of establishing a kingdom, once the opposite cities find that an overlapping of domain is unavoidable, they will inevitably engage in a conflict.

Sometimes for a temporary benefit a bunch of kings may ally into a party. It is certain that one king can only take part in one party at one time, and the cities following a king will correspondingly support the party he joins.

Rather than fighting each other fiercely, cities supporting the same party cooperate and strive to establish a kingdom of the whole party.

Such kingdoms follow the similar rules:

1. The kingdom is a rectangle on the map.
2. Every city on the Gridland supporting that party should be inside the kingdom.

And the fighting trigger with other kingdoms is exactly the same as the above one.

A established kingdom is called stable if and only if:

1. No city in the kingdom has to fight with another kingdom..
2. If the kingdom stands for a party, the party must never have more than R kings.

Now, your job is to find the largest possible stable kingdom. By `largest' we mean the one with the largest number of cities.

 

Input

 Input contains multiple cases. Each case consists of:

Line 1:
Two integers: N and M (1 ≤ N, M ≤ 1000)
Line 2:
Two integers: K and R (1 ≤ K ≤ 1000)
Line 3..N + 2 :
M integers each line. the j -th integer on the i + 2 -th line will be between 1 and k , indicating which king that city supports.

Output

 For each case, output one line containing an integer that is the area of the largest possible stable kingdom.

Sample Input

3 3 5 3 4 1 1 2 3 1 2 2 5 10 10 13 5 1 1 1 4 4 4 4 4 3 11 1 1 1 4 4 4 4 4 11 3 1 1 1 2 2 2 2 2 3 3 1 1 1 2 2 2 2 2 3 11 1 1 1 2 2 2 2 2 11 11 1 1 1 2 2 2 2 2 3 11 1 1 1 6 6 6 7 7 3 3 1 1 1 6 6 6 7 7 3 3 5 5 5 5 10 13 8 12 9 9 5 5 5 5 13 10 8 8 9 9 2 2 2 1 2 2 2 1

Sample Output

1 64 
0
// Problem#: 1914// Submission#: 3591445// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int MAXN = 1001;const int MAXK = 1001;int n, m, k, r;short matrix[MAXN][MAXN];int s0_input() {    if (scanf("%d%d\n", &n, &m) != 2) return 0;    scanf("%d%d\n", &k, &r);    char temp[10000];    for (int i = 0; i < n; i++) {        short j = 0, sum = 0;        gets(temp);        for (int k = 0; temp[k] != '\0'; k++) {            if (temp[k] == ' ') {                matrix[i][j++] = sum;                sum = 0;            } else sum = sum * 10 + temp[k] - '0';        }        matrix[i][j++] = sum;    }    return 1;}int x1[MAXK], x2[MAXK], y1[MAXK], y2[MAXK];void s1_king_pos() {    for (int i = 1; i <= k; i++) {        x1[i] = y1[i] = MAXN;        x2[i] = y2[i] = -1;    }    for (int i = 0; i < n; i++) {        for (int j = 0; j < m; j++) {            int king = matrix[i][j];            x1[king] = min(x1[king], i);            x2[king] = max(x2[king], i);            y1[king] = min(y1[king], j);            y2[king] = max(y2[king], j);        }    }}struct Border {    int id, type;    int row() const {        if (type == 1) return x1[id];        return x2[id] + 1;    }    int col() const {        if (type == 1) return y1[id];        return y2[id] + 1;    }    Border(int i = 0, int j = 0) {        id = i;        type = j;    }}border[MAXK * 2];bool cmp_row(const Border & a, const Border & b) {    return a.row() < b.row();}bool cmp_col(const Border & a, const Border & b) {    return a.col() < b.col();}int left[MAXN + 1][MAXN + 1];int up[MAXN + 1][MAXN + 1];void s2_outline() {    for (int i = 0; i < k; i++) {        border[i] = Border(i + 1, 1);        border[i + k] = Border(i + 1, -1);    }    sort(border, border + k * 2, cmp_col);    for (int i = 1; i < n; i++) {        int c = 0, p = 0;        left[i][0] = 0;        for (int j = 0; j < m; j++) {            for (; p < k * 2 && border[p].col() == j; p++) {                int u = border[p].id;                if (x1[u] < i && x2[u] >= i) c += border[p].type;            }            if (c > 0) left[i][j + 1] = 0;            else left[i][j + 1] = left[i][j] + 1;        }    }    for (int i = 0; i <= m; i++) left[0][i] = left[n][i] = i;    sort(border, border + k * 2, cmp_row);    for (int i = 1; i < m; i++) {        int c = 0, p = 0;        up[0][i] = 0;        for (int j = 0; j < n; j++) {            for (; p < k * 2 && border[p].row() == j; p++) {                int u = border[p].id;                if (y1[u] < i && y2[u] >= i) c += border[p].type;            }            if (c > 0) up[j + 1][i] = 0;            else up[j + 1][i] = up[j][i] + 1;        }    }    for (int i = 0; i <= n; i++) up[i][0] = up[i][m] = i;}int counter[MAXN][MAXN];void s3_pre_count() {    for (int i = 0; i < n; i++) memset(counter[i], 0, sizeof(int) * m);    for (int i = 1; i <= k; i++) {        if (x2[i] >= 0) counter[x1[i]][y1[i]]++;    }    for (int i = 0; i < n; i++) {        for (int j = 0; j < m; j++) {            int & t = counter[i][j];            if (i > 0) t += counter[i - 1][j];            if (j > 0) t += counter[i][j - 1];            if (i > 0 && j > 0) t -= counter[i - 1][j - 1];        }    }}int s4_enumerate() {    int ret = 0;    for (int i = 1; i <= k; i++) {        for (int j = 1; j <= k; j++) {            if (x2[i] < 0 || x2[j] < 0) continue;            int height = x2[j] - x1[i] + 1;            int width = y2[j] - y1[i] + 1;            if (height <= 0 || width <= 0) continue;            if (up[x2[j] + 1][y1[i]] < height) continue;            if (up[x2[j] + 1][y2[j] + 1] < height) continue;            if (left[x1[i]][y2[j] + 1] < width) continue;            if (left[x2[j] + 1][y2[j] + 1] < width) continue;            int c = counter[x2[j]][y2[j]];            if (x1[i] > 0) c -= counter[x1[i] - 1][y2[j]];            if (y1[i] > 0) c -= counter[x2[j]][y1[i] - 1];            if (x1[i] > 0 && y1[i] > 0) c += counter[x1[i] - 1][y1[i] - 1];            if (c > r) continue;            ret = max(ret, height * width);        }    }    return ret;}int main() {    while (s0_input()) {        s1_king_pos();        s2_outline();        s3_pre_count();        printf("%d\n", s4_enumerate());    }    return 0;}