Now a spy is besieged in a maze. He knows that there is a telegraph transmitter in the maze somewhere. So he must get there to use the telegraph transmitter to ask for help.
But the mission is very dangerous, because there are some mines in the map (see the figure above). When the spy steps on it, it will immediately explode and the mission will fail. The spy doesn‘t know where the mines are, thus he will use his best strategy to get to the destination (he will always go along the shortest path). There may be several shortest paths for the spy to choose, and the probability to choose each path from the start point to the destination is the same.
Now your task is to write a program to find the probability that the spy can get to the telegraph transmitter.
Input
The input file begins with an integer T, indicating the number of test cases. Each test case begins with two integers N, M, indicating the height and width of the maze. (N <= 10, M <= 10) In the following N lines, each line contains M characters describing the map. There is one blank line after each map. Spaces denotes empty square, ‘#‘ denotes a wall, ‘S‘ denotes the spy, ‘M‘ denotes a mine, and ‘T‘ denotes the telegraph transmitter. It‘s guaranteed that the four sides of the map are all walls.
Output
For each maze, first output the number of the test case (`Mission #1:‘, ` Mission #2:‘, etc.) in a line of its own.
If it is possible for the spy to get to the telegraph transmitter, print a line containing the probability that the spy can get to the telegraph transmitter, exact to two digit to the right of the decimal point. Adhere to the output format shown in the sample below.
If the spy can‘t get to the destination, output a line containing the statement `Mission Impossible.‘
Output a blank line after each test case.
Sample Input
2 6 10 ########## # M T # # ### # # ### # # S # ##########
6 10 ########## # M T # # ### # # ### # # S # ##########
Sample Output
Mission #1: The probability for the spy to get to the telegraph transmitter is 50.00%.
Mission #2: Mission Impossible.
Author: YE, Kai Source: ZOJ Monthly, February 2004
题意：间谍只走最短路，而这些路上可能埋有炸弹，请问间谍安全到达目的地的可能性多少。
题解：BFS记录最短路径个数以及这些路径中踩到炸弹的路径的个数，最后换算个百分比就行了。
注意需要延迟标记，还有图中有空格，读入不能用scanf。
1 #include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<map> 9 10 #define N 101 11 #define M 15 12 #define mod 1000000007 13 #define mod2 100000000 14 #define ll long long 15 #define maxi(a,b) (a)>(b)? (a) : (b) 16 #define mini(a,b) (a)<(b)? (a) : (b) 17 18 using namespace std; 19 20 int T; 21 char s[N][N]; 22 int vis[N][N]; 23 int cc,tot; 24 int n,m; 25 int dirx[]={1,-1,0,0}; 26 int diry[]={0,0,1,-1}; 27 28 typedef struct 29 { 30 int x; 31 int y; 32 int now; 33 int boom; 34 }PP; 35 36 PP start,end; 37 38 int ok(PP next) 39 { 40 if(next.x>=0 && next.x<n && next.y>=0 && next.y<m && vis[next.x][next.y]==0 && s[next.x][next.y]!=‘#‘){ 41 if(s[next.x][next.y]==‘M‘) 42 next.boom=1; 43 return next.boom; 44 } 45 return 0; 46 } 47 48 void BFS() 49 { 50 queue<PP> q; 51 q.push(start); 52 int first=-1; 53 PP te,next; 54 int i; 55 int re; 56 while(q.size()!=0) 57 { 58 te=q.front(); 59 vis[te.x][te.y]=1; 60 q.pop(); 61 // printf(" %d %d\n",te.x,te.y); 62 for(i=0;i<4;i++){ 63 next.x=te.x+dirx[i]; 64 next.y=te.y+diry[i]; 65 next.now=te.now+1; 66 next.boom=te.boom; 67 re=ok(next); 68 next.boom=re; 69 if(re==0) continue; 70 if(next.x==end.x && next.y==end.y) 71 { 72 // printf(" %d %d first=%d now=%d %d\n",next.x,next.y,first,next.now,next.boom); 73 if(first==-1) first=next.now; 74 else{ 75 if(next.now!=first) return; 76 } 77 tot++; 78 if(next.boom==2) cc++; 79 } 80 else{ 81 q.push(next); 82 } 83 } 84 } 85 } 86 87 int main() 88 { 89 int i,j; 90 //freopen("data.in","r",stdin); 91 scanf("%d",&T); 92 getchar(); 93 for(int cnt=1;cnt<=T;cnt++) 94 //while(T--) 95 //while(scanf("%d%d",&n,&q)!=EOF) 96 { 97 //q.clear(); 98 cc=tot=0; 99 memset(vis,0,sizeof(vis)); 100 scanf("%d%d",&n,&m); 101 getchar(); 102 for(i=0;i<n;i++){ 103 gets(s[i]); 104 } 105 for(i=0;i<n;i++){ 106 for(j=0;j<m;j++){ 107 if(s[i][j]==‘S‘){ 108 start.x=i; 109 start.y=j; 110 start.boom=2; 111 start.now=0; 112 } 113 if(s[i][j]==‘T‘){ 114 end.x=i; 115 end.y=j; 116 } 117 } 118 } 119 //printf("%d %d %d %d\n",start.x,start.y,end.x,end.y); 120 BFS(); 121 printf("Mission #%d:\n",cnt); 122 if(cc==0){ 123 printf("Mission Impossible.\n"); 124 } 125 else{ 126 printf("The probability for the spy to get to the telegraph transmitter is %.2f%%.\n",(1.0)*100*cc/tot); 127 } 128 printf("\n"); 129 130 } 131 132 return 0; 133 }
zoj 2081 BFS 延迟标记 读入问题,布布扣,bubuko.com
原文：http://www.cnblogs.com/njczy2010/p/3917599.html