试题描述：
已经连续下了几天雨，却还是没有停的样子。土豪CCY刚从外地赚完1e元回来，知道不久除了自己别墅，其他的地方都将会被洪水淹没。
CCY所在的城市可以用一个N*M(N,M<=50)的地图表示，地图上有五种符号：“. * X D S”。其中“X”表示石头，水和人都不能从上面经过。“.”表示平原，CCY和洪水都可以经过。“*”表示洪水开始地方（可能有多个地方开始发生洪水）。“D”表示CCY的别墅。“S”表示CCY现在的位置。
CCY每分钟可以向相邻位置移动，而洪水将会在CCY移动之后把相邻的没有的土地淹没（从已淹没的土地）。
求CCY回到别墅的最少时间。如果聪哥回不了家，就很可能会被淹死，那么他就要膜拜黄金大神涨RP来呼叫直升飞机，所以输出“ORZ hzwer!!!”。
输入：
第一行：N和M
接下来N行，每行M个字符，表示地图。
输出：
若能回家，输出一个整数，表示最少回家步数；否则输出“ORZ hzwer!!!”。
输入示例：
3 3
D.*
...
.S.
输出示例：
3
良心输入示例：
50 50 ...SXX...XX...X...XX.XX....X.......X....X.X....*X. .XX.X..XX.X.X..X.X.X...XX.X.....X....X...X.XX....X ..X..X....X.X.X.X.XX....XX...X..XX....X...XXX...XX X.....X..XXX.XX..X.X.X........XXX....X.X..XX..X... X..X.....X......X.......X..X......X.....X.X..X.... ......XXX..X..XX.X.......X..X......X.X.XXX.X..X... ...X...XX.X..X.........X..XX..XX.X..X.X....X..X..X XX.........XX.XXX.XXX...X..X..XXX.XX.X..X...X.X.X. .....X...X..X.X.X.....X.X.X...XX........XX..X..X.. XXXX..XX..X..X..X..XXX..X...X..X.X...X..XX.XX.X... ......X..X....X.........XX..X..X........X.XX...X.X X..XX...X........X........XXXX.XX.........X.X.XXX. .....XXX...X.XXX....X...XX........X...XXXX...X.... .X.XX....XXX...X...X...X.X...XXX...X...X..XX.XX... ...XXX....XXX.....X.X...XX...X.X.XXX.X.XX..X...... .......X..XX.X.X..........X.XX..X..X.XX.XXXX...XX. X...X.X.X.XXX.X..X.XXX..X...X....X.X..X...X..XX..X ..X..X.XX..X.XXX..X...X..X..X..X.X....X..........X XX....X.X.X.XX.....X..XXXX.XX..X....X.....X.....X. .XX.....X.X..X..X.X...X......X....X..X.XX.XX.XX.X. .....X.........X..X.X...X....XX.X...X..X..XX.XX... .X........X..XX..X...X....XX.XXX.XX.X..X...X...XXX ...X.X.XX...X......XX.X.XXX..X.X.XX....XXX.X...X.. ....X.....XX..XX.X.X...X..X.X....X.....XX.XX..XXX. ....X........XXXXXX.........XX..X....XX.XXXX...... ..XX.XXX.X...X.X.X.X..XXXX....X.X.X...XX.X..X..X.. XX..XX.X...X....XX....X.......X..X.X......XXXXX... XX..XX.XX.X..X...XX....XX.X..X...X.X....X......... ..X.....X...X..X.........X......X...XX.XX...X..... X.X..XX.X.X.X.XXXX.X...X..X...XX..X.....X.X.XXX... ..X....X.X.X......X..X...XX..X...X.X.XXX.X....XX.X X.....X.X...XXXXX.....X..XX...X....X...XXX....X... .XXX...XXXX.X....X...XX.XX..X.X........X...X....X. ..XXX.XX....X.XX.X..X...X.X.X.....X...X....X.XXXXX .XX...X.X....XXX......X.....X....X.X..X..X....XX.X XX..XX.X....X...XXX....X..XXX.XXX.X.X....X.X...... XX....X.X....X..X.XX...X..X.X....X..XX...XX...X... X.X...X...X..X....X.X.........X...X...X....X..X... .X...X...X.X....X.....XX......XXX..XX.XX.X.....X.. ..X.X..XX.X..X..........X.X.XX..X..X.X.XXXX....... ..X...XX.X.X..XX..XXXXX...X........XX...X.X..X..XX .....X..XXX....X.XX....XX.XX..XX.X....XX.XX..X.XXX X..X.XX.X.X.......X.X.X..X......X..XX....X.XXX.XXX X..XXXX.........X..X..X............X.X..X..X...X.. X...XX.XX...X........X....X....X.X...X..XD...X..X. ......X.......XXXX..XXX.....X.X.X.X..........X.... X.....XX.XX.XXX.XX.X.X..X.X.......XXXX.....X...XXX .X..XX..X.....XX...XX..X.....XX...X..X....X..X..XX X....XX....XX.....XX.X.XXX....X.......X..X....X..X X.....XX..XX.X.....XX....X..X..X.X.XX.......X..X..
良心输出示例：
86
解题思路：
这是一道神奇的题目，我用了一天。。。。。
首先，我写了一个DFS+BFS，然后结果错误。
接着，改成预处理，创建一个step数组，step[i][j]表示洪水第几步漫道这里。
如：
3 3
D.*
...
.S.
的step是：
-1 1 0
3 2 1
4 3 2
这需要一个BFS
读入·时每遇到一个*就把它丢到队列里面，
然后正常BFS。
预处理以后，就再正常地BFS
然后就好了。
#include <iostream> #include <algorithm> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <queue> using namespace std; int map[51][51]; int step[51][51]; int book[51][51]; int startx, starty; int endx,endy; queue <int> Qx; queue <int> Qy; int n, m; int ans = 0; int wnum = 0; void getstep() { while(!Qx.empty()) { int kx = Qx.front(); int ky = Qy.front(); Qx.pop(); Qy.pop(); if(map[kx-1][ky]==0 && step[kx-1][ky]>step[kx][ky]+1 && kx-1>0) { Qx.push(kx-1); Qy.push(ky); step[kx-1][ky]=step[kx][ky]+1; } if(map[kx+1][ky]==0 && step[kx+1][ky]>step[kx][ky]+1 && kx+1<=n) { Qx.push(kx+1); Qy.push(ky); step[kx+1][ky]=step[kx][ky]+1; } if(map[kx][ky-1]==0 && step[kx][ky-1]>step[kx][ky]+1 && ky-1>0) { Qx.push(kx); Qy.push(ky-1); step[kx][ky-1]=step[kx][ky]+1; } if(map[kx][ky+1]==0 && step[kx][ky+1]>step[kx][ky]+1 && ky+1<=m) { Qx.push(kx); Qy.push(ky+1); step[kx][ky+1]=step[kx][ky]+1; } } } int main() { memset(book,88,sizeof(book)); memset(step,88,sizeof(step)); cin>>n>>m; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { char c; cin >> c; if (c == 'S') { startx = i; starty = j; } if (c == '*') { Qx.push(i); Qy.push(j); step[i][j]=0; } if (c == 'D') { endx=i; endy=j; map[i][j]=2; } if (c == 'X') map[i][j] = -1; } getstep(); queue <int> qx; queue <int> qy; qx.push(startx); qy.push(starty); book[startx][starty]=0; while(!qx.empty()) { int kx=qx.front(); int ky=qy.front(); qx.pop(); qy.pop(); if(map[kx][ky]==2) break; if((map[kx-1][ky]==0 || map[kx-1][ky]==2) && book[kx-1][ky]>book[kx][ky]+1 && step[kx-1][ky]>book[kx][ky]+1 && kx-1>=1) { qx.push(kx-1); qy.push(ky); book[kx-1][ky]=book[kx][ky]+1; } if((map[kx+1][ky]==0 || map[kx+1][ky]==2) && book[kx+1][ky]>book[kx][ky]+1 && step[kx+1][ky]>book[kx][ky]+1 && kx+1<=n) { qx.push(kx+1); qy.push(ky); book[kx+1][ky]=book[kx][ky]+1; } if((map[kx][ky+1]==0 || map[kx][ky+1]==2) && book[kx][ky+1]>book[kx][ky]+1 && step[kx][ky+1]>book[kx][ky]+1 && ky+1<=m) { qx.push(kx); qy.push(ky+1); book[kx][ky+1]=book[kx][ky]+1; } if((map[kx][ky-1]==0 || map[kx][ky-1]==2) && book[kx][ky-1]>book[kx][ky]+1 && step[kx][ky-1]>book[kx][ky]+1 && ky-1>=1) { qx.push(kx); qy.push(ky-1); book[kx][ky-1]=book[kx][ky]+1; } } if(book[endx][endy]>=1000000000) cout<<"ORZ hzwer!!!"; else cout<<book[endx][endy]; }