zoj 2081 Mission Impossible

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2081

 

纠结一个晚上的题目..竟然同时使用了BFS和DFS.菜鸟花了一个晚上才醒悟过来..

 

好吧..终于了解了DFS和BFS之间的区别了..oh yeah~~

 

/* * 2081.cpp * * Created on: Apr 30, 2010 * Author: wyy * */#define MaxVertexs 10#include<cstdio>#include<queue>using namespace std;struct point{int x, y;};int dx[4] = {0, 1, 0, -1};int dy[4] = {1, 0, -1, 0};point S, T, B[80];int N, M, repeat, MinDis, Count, Sum, Bn;int dis[MaxVertexs][MaxVertexs];char graph[MaxVertexs][MaxVertexs];bool visited[MaxVertexs][MaxVertexs];void Read(){Bn = Count = Sum = 0;scanf("%d%d", &N, &M);getchar();for(int i = 0; i < N; ++i){gets(graph[i]);for(int j = 0; j < M; ++j){dis[i][j] = 0;if(graph[i][j] == 'S'){S.x = i;S.y = j;}else if(graph[i][j] == 'T'){T.x = i;T.y = j;}else if(graph[i][j] == 'M'){++Bn;B[Bn].x = i;B[Bn].y = j;}}}}void Init(){for(int i = 0; i < N; ++i)for(int j = 0; j < M; ++j)visited[i][j] = false;visited[S.x][S.y] = true;}bool IsValid(const point &p){return (p.x > 0 && p.y > 0 && p.x < N - 1 && p.y < M - 1 && graph[p.x][p.y] != '#' && !visited[p.x][p.y]);}void BFS(){point m, n;queue<point> Q;Q.push(S);while(!Q.empty()){m = Q.front();Q.pop();for(int i = 0; i < 4; ++i){n.x = m.x + dx[i];n.y = m.y + dy[i];if(IsValid(n)){dis[n.x][n.y] = dis[m.x][m.y] + 1;visited[n.x][n.y] = true;Q.push(n);}}}MinDis = dis[T.x][T.y];}void DFS(int size, int c, point p){if(size == MinDis && p.x == T.x && p.y == T.y){++Sum;if(c > 0)++Count;}else if(size < MinDis){point m;for(int i = 1; i <= Bn; ++i)if(p.x == B[i].x && p.y == B[i].y){++c;break;}for(int i = 0; i < 4; ++i){m.x = p.x + dx[i];m.y = p.y + dy[i];if(IsValid(m)){visited[m.x][m.y] = true;DFS(size + 1, c, m);visited[m.x][m.y] = false;}}}}void Solve(int k){Read();Init();BFS();Init();DFS(0, 0, S);printf("Mission #%d:/n", k + 1);if(Sum > Count)printf("The probability for the spy to get to the telegraph transmitter is %.2lf%%./n/n", 100.0 * (Sum - Count) / Sum);elseprintf("Mission Impossible./n/n");}int main(){//freopen("input.txt", "r", stdin);scanf("%d", &repeat);for(int i = 0; i < repeat; ++i)Solve(i);//fclose(stdin);return 0;}