poj 3467 Cross Counting

题意：给出一个M*N 的矩阵，矩阵中每个元素都有一种颜色(1 ≤ N, M, C ≤ 100),10000条查询，更改某个元素的颜色或查询某种颜色的corss有多少个。

We say there exists a cross of size k centered at the cell (x,y)iff all cells lying in the x-th row or the y-th column and within a distance of k from (x,y) share the same color. if two crosses have the same center but different sizes we consider they are distinct

解法：由于每种颜色之间互不相关，因此可以分颜色考虑，将矩阵变为100个01矩阵，这样可以统计出初始时每种颜色有多少个cross。

当某个元素(i,j)发生改变时，只影响i行或j列为中心的cross，因此只需逐个考虑每个中心的cross的变化，每次查询只需更改200的元素的cross数目。怎么快速统计出以每个元素为中心的cross数目呢?根据定义，只需记录01矩阵中每个元素向上下左右四个方向连续1的个数，取最小值即可，当一个元素从1变为0或从0变为1，只需改变同行元素的左右方向连续1的个数和同列元素的上下方向连续1的个数，并记录cross数目增加减少的数量即可。

貌似线段树也可以维护，没想到这么维护，所以就暂且认为复杂度比这个高了（囧）。。。

import java.util.Scanner;public class CountCross3467 {int maxn = 110;class Matrix {int map[][] = new int[maxn][maxn];int left[][] = new int[maxn][maxn], right[][] = new int[maxn][maxn],up[][] = new int[maxn][maxn], down[][] = new int[maxn][maxn];int sum, n, m;        Matrix(int n,int m){        this.n=n;        this.m=m;        }void init(int i, int j) {map[i][j] = 1;left[i][j] = right[i][j] = j;up[i][j] = down[i][j] = i;}void count() {for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++)if (map[i][j] == 1 && map[i][j - 1] == 1)left[i][j] = left[i][j - 1];for (int j = m; j > 0; j--)if (map[i][j] == 1 && map[i][j + 1] == 1)right[i][j] = right[i][j + 1];}for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++)if (map[j][i] == 1 && map[j - 1][i] == 1)up[j][i] = up[j - 1][i];for (int j = n; j > 0; j--)if (map[j][i] == 1 && map[j + 1][i] == 1)down[j][i] = down[j + 1][i];}sum = 0;for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++) sum += cross(i, j);}int cross(int i, int j) {if(map[i][j]==0)return 0;int a = Math.min(i - up[i][j], down[i][j] - i);int b = Math.min(j-left[i][j], right[i][j] - j);return Math.min(a, b);}void update1(int a, int b) {// 0-->1map[a][b] = 1;left[a][b] = b;if (map[a][b - 1] == 1)left[a][b] = left[a][b - 1];right[a][b] = b;if (map[a][b + 1] == 1)right[a][b] = right[a][b + 1];up[a][b] = a;if (map[a - 1][b] == 1)up[a][b] = up[a - 1][b];down[a][b] = a;if (map[a + 1][b] == 1)down[a][b] = down[a + 1][b];sum+=cross(a,b);for (int i = 1; i < b; i++) {if (right[a][i] != b - 1)continue;int temp = cross(a, i);right[a][i] = right[a][b];sum += cross(a, i) - temp;}for (int i = b + 1; i <= m; i++) {if (left[a][i] != b + 1)continue;int temp = cross(a, i);left[a][i] = left[a][b];sum += cross(a, i) - temp;}for(int i=1;i<a;i++){if(down[i][b]!=a-1)continue;int temp=cross(i,b);down[i][b]=down[a][b];sum += cross(i,b) - temp;}for(int i=a+1;i<=n;i++){if(up[i][b]!=a+1)continue;int temp=cross(i,b);up[i][b]=up[a][b];sum += cross(i,b) - temp;}}void update0(int a, int b) {// 1-->0sum-=cross(a,b);map[a][b] = 0;left[a][b] =right[a][b]=up[a][b]=down[a][b]=0;for (int i = 1; i < b; i++) {if (map[a][i]==0||right[a][i]<b)continue;int temp = cross(a, i);right[a][i] =b-1;sum -= temp-cross(a, i);}for (int i = b + 1; i <= m; i++) {if (map[a][i]==0||left[a][i]>b)continue;int temp = cross(a, i);left[a][i] = b+1;sum -= temp-cross(a, i);}for(int i=1;i<a;i++){if(map[i][b]==0||down[i][b]<a)continue;int temp=cross(i,b);down[i][b]=a-1;sum-= temp-cross(i,b);}for(int i=a+1;i<=n;i++){if(map[i][b]==0||up[i][b]>a)continue;int temp=cross(i,b);up[i][b]=a+1;sum -= temp-cross(i,b);}}}Scanner scan=new Scanner(System.in);void run(){int n=scan.nextInt();int m=scan.nextInt();int c=scan.nextInt();int q=scan.nextInt();int map[][]=new int[n+1][m+1];Matrix mat[]=new Matrix[c+1];for(int i=1;i<=c;i++)mat[i]=new Matrix(n,m);for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){map[i][j]=scan.nextInt();mat[map[i][j]].init(i, j);}for(int i=1;i<=c;i++)mat[i].count();while(q-->0){String s=scan.next();if(s.charAt(0)=='Q'){int v=scan.nextInt();System.out.println(mat[v].sum);}else{int i=scan.nextInt();int j=scan.nextInt();int v=scan.nextInt();int t=map[i][j];map[i][j]=v;mat[t].update0(i, j);mat[v].update1(i, j);}}}public static void main(String[] args) {new CountCross3467().run();}}