BFS 宽搜 B题 - Igor In the Museum

题意是一个人去参观画室，每一块空地和一个墙之间都有一幅画，给你这个人在空地的位置，他可以到处走动，问此人最多能看到多少幅画。我说说我解这个题的过程:起初没考虑时间复杂度，直接用BFS搜索,然后TLE了，然后一看如果有10W个指令， 不超时才怪。然后想到先保存查询的状态然后查询结果就行。然后又Runtime error in test 11. 这是什么鬼错误，原来是内存溢出，检查一下原来分配点空间的时候没分配够。最后AC了。总结了经验:出现错误有两种可能，一种是思路错了，这是致命伤，还有一种可能是非常细节的地方没考虑到，或者细微的地方考虑到了，但是打程序的时候给打错了，例如本来我要赋值呢 a = b;我写成了a == b;然后感觉对，却输不出正确的结果。很伤心的错误，再犯就剁手。DescriptionIgor is in the museum and he wants to see as many pictures as possible.Museum can be represented as a rectangular field of n × m cells. Each cell is either empty or impassable. Empty cells are marked with '.', impassable cells are markedwith '*'. Every two adjacent cells of different types (one empty and one impassable) are divided by a wall containing one picture.At the beginning Igor is in some empty cell. At every moment he can move to any empty cell that share a side with the current one.For several starting positions you should calculate the maximum number of pictures that Igor can see. Igor is able to see the picture only if he is in the cell adjacent to the wall with this picture. Igor have a lot of time, so he will examine every picturehe can see.InputFirst line of the input contains three integers n,m and k (3 ≤ n, m ≤ 1000, 1 ≤ k ≤ min(n·m, 100 000)) — the museum dimensions and the number ofstarting positions to process.Each of the next n lines containsm symbols '.', '*' — the description of the museum. It is guaranteed that all border cells are impassable, so Igor can't go out from the museum.Each of the last k lines contains two integersx and y (1 ≤ x ≤ n, 1 ≤ y ≤ m) — the row and the column of one of Igor's starting positions respectively. Rows arenumbered from top to bottom, columns — from left to right. It is guaranteed that all starting positions are empty cells.OutputPrint k integers — the maximum number of pictures, that Igor can see if he starts in corresponding position.Sample InputInput

5 6 3
******
*..*.*
******
*....*
******
2 2
2 5
4 3

Output

6
4
10

Input

4 4 1
****
*..*
*.**
****
3 2

Output

8

AC代码如下：[code]#include <iostream>#include <cstdio>#include <cstdlib>#include <queue>#include <cstring>using namespace std;//#define LOCAL#define MAX_N 1005#define INF 100000000typedef pair<int, int> P;//点坐标int path[1000005][3];//最多有十万个点int px=0, fir=0;char mus[MAX_N][MAX_N];//存放博物馆int n, m, k;//n是高,m是宽, k是起始点int bj[MAX_N][MAX_N];//标记点是否走过int dir[4][2] = {1,0, 0,1, -1,0, 0,-1};int bfs(int x, int y){int ans = 0;//建立队列queue<P> que;que.push(P(x,y));//起始点如队列fir = px;//记录每次df的第一个path[px][0] = x;//入队的点都记录path[px][1] = y;px++;bj[x][y] = 1;//标记起始点已经走过//进行每个点的搜索while(!que.empty()){//取出队列中的点P p = que.front();que.pop();//别忘了出来for(int i=0; i<4; i++){//四个方向int nx = p.first + dir[i][1], ny = p.second + dir[i][0];if(mus[nx][ny] == '.' && bj[nx][ny] == 0){//只有.时能走que.push(P(nx, ny));//点入队列path[px][0] = nx, path[px][1] = ny;//入队的点都标记px++;bj[nx][ny] = 1;//标记走过}elseif(mus[nx][ny] == '*'){//是墙的时候ans++;}}}return ans;}int main(){#ifdef LOCALfreopen("b:\\data.in.txt", "r", stdin);#endifwhile(~scanf("%d%d%d", &n, &m, &k)){px=0, fir=0;memset(bj, 0, sizeof(bj));//重置0//输入迷宫for(int i=0; i<n; i++)for(int j=0; j<m; j++){cin>>mus[i][j];//            cout<<mus[i][j]<<" ";//            if(j == m-1) cout<<endl;}//不能直接对K个点进行操作,否则会超时for(int i=0; i<n; i++)for(int j=0; j<m; j++){//对每一个能走的快操作if(mus[i][j] == '.' && bj[i][j] == 0){//没有检测过再dfsint ret = bfs(i, j);for(int ed=fir; ed<px; ed++){path[ed][2] =  ret;//标记好点对应的结果}}}for(int i=0; i<px; i++){bj[ path[i][0] ][ path[i][1] ] = path[i][2];//用标记数组代替结果存储数组，实现内存节省}//        for(int i=0; i<px; i++)//            cout<<path[i][0]<<"-"<<path[i][1]<<"-"<<path[i][2]<<endl;int a, b;while(k--){scanf("%d%d", &a, &b);printf("%d\n", bj[a-1][b-1]);}}return 0;}

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