先预处理出每一个联通块的答案,然后查询时就可以O(1)出答案;
/* *********************************************** Author :xdlove Created Time :2015年11月15日 星期日 16时47分40秒 File Name :xdlove/codeforces/Educational_Codeforces_Round_1/D/D.cpp ************************************************ */ #pragma comment(linker, "/STACK:1024000000,1024000000") #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <memory.h> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; #define REP_ab(i,a,b) for(int i = a; i <= b; i++) #define REP(i, n) for(int i = 0; i < n; i++) #define REP_1(i,n) for(int i = 1; i <= n; i++) #define DEP(i,n) for(int i = n - 1; i >= 0; i--) #define DEP_N(i,n) for(int i = n; i >= 1; i--) #define CPY(A,B) memcpy(A,B,sizeof(B)) #define MEM(A) memset(A,0,sizeof(A)) #define MEM_1(A) memset(A,-1,sizeof(A)) #define MEM_INF(A) memset(A,0x3f,sizeof(A)) #define MEM_INFLL(A) memset(A,0x3f3f,sizeof(A)) #define mid (((l + r) >> 1)) #define lson l, mid, u << 1 #define rson mid + 1, r, u << 1 | 1 #define ls (u << 1) #define rs (u << 1 | 1) typedef long long ll; typedef unsigned long long ull; const int INF = 0x3f3f3f3f; const ll INFLL = 0x3f3f3f3f3f3f3f3f; const int MAXN = 1e3 + 5; const int MAXM = MAXN; const int mod = 1e9 + 7; char s[MAXN][MAXN]; int n,m,k; int vis[MAXN][MAXN]; int res[MAXN * MAXN]; int dx[] = { 0,0,-1,1 }; int dy[] = { -1,1,0,0 }; int id; bool check(int x,int y) { if(x < 0 || y < 0 || x >= n || y >= m || vis[x][y]) return false; return true; } int dfs(int x,int y) { vis[x][y] = id; int cnt = 0; for(int i = 0; i < 4; i++) { int nx = dx[i] + x; int ny = dy[i] + y; if(check(nx,ny)) { if(s[nx][ny] != '.') cnt++; else cnt += dfs(nx,ny); } } return cnt; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); cin>>n>>m>>k; REP(i,n) scanf("%s",s[i]); id = 1; MEM(vis); REP(i,n) { REP(j,m) { if(s[i][j] == '.' && !vis[i][j]) { res[id] = dfs(i,j); id++; } } } while(k--) { int x,y; scanf("%d %d",&x,&y); x--,y--; printf("%d\n",res[vis[x][y]]); } return 0; }