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Battle City
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7579 Accepted: 2544 Description
Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now.
What we are discussing is a simple edition of this game. Given a map that consists of empty spaces, rivers, steel walls and brick walls only. Your task is to get a bonus as soon as possible suppose that no enemies will disturb you (See the following picture).
Your tank can't move through rivers or walls, but it can destroy brick walls by shooting. A brick wall will be turned into empty spaces when you hit it, however, if your shot hit a steel wall, there will be no damage to the wall. In each of your turns, you can choose to move to a neighboring (4 directions, not 8) empty space, or shoot in one of the four directions without a move. The shot will go ahead in that direction, until it go out of the map or hit a wall. If the shot hits a brick wall, the wall will disappear (i.e., in this turn). Well, given the description of a map, the positions of your tank and the target, how many turns will you take at least to arrive there?Input
The input consists of several test cases. The first line of each test case contains two integers M and N (2 <= M, N <= 300). Each of the following M lines contains N uppercase letters, each of which is one of 'Y' (you), 'T' (target), 'S' (steel wall), 'B' (brick wall), 'R' (river) and 'E' (empty space). Both 'Y' and 'T' appear only once. A test case of M = N = 0 indicates the end of input, and should not be processed.Output
For each test case, please output the turns you take at least in a separate line. If you can't arrive at the target, output "-1" instead.Sample Input
3 4 YBEB EERE SSTE 0 0
Sample Output
8
Y是自己 B是砖墙（可以打破） S是铁墙（无法通过） R是河流（无法通过） T是敌人 当遇到B花费加2当遇到E花费加1
/* 题意：小时候玩的坦克大战一样，字符Y是自己 B是砖墙可以打破 但是打破需要花费两点能量，S是铁墙无法通过，R是河流无法通过 T是敌人，E是路通过需要花费一点能量，问杀死敌人需要花费多少能量 （敌人不会动）并输出花费的能量 如果不能杀死敌人输出-1 题解： 开一个优先队列，用来记录坐标和花费的能量找到自己的位置坐标，然后开始向四个方向搜索， 每次经过空地将能量加1，经过砖墙将能量加2 */ #include<stdio.h> #include<string.h> #include<queue> using namespace std; #define MAX 1100 char map[MAX][MAX]; int vis[MAX][MAX]; int x1,x2,y1,y2; int n,m; struct node//建立结构体优先队列 { int a; int b; int cost; friend bool operator < (node a,node b) { return a.cost>b.cost;//表示花费能量少的优先出队 } }; void bfs() { int i,j,t; int move[4][2]={0,1,0,-1,-1,0,1,0}; node beg,end; priority_queue<node>q; beg.a=x1;//起点坐标 beg.b=y1; beg.cost=0; q.push(beg);//将起点坐标入队 vis[x1][y1]=1;//标记走过的点，防止重复搜索 while(!q.empty())//如果队列不为空 { end=q.top();//去队顶元素 q.pop();//删除队顶元素 if(end.a==x2&&end.b==y2)//如果找到终点即找到敌人 { //敌人坐标为终点坐标 printf("%d ",end.cost); return ; } for(i=0;i<4;i++) { beg.a=end.a+move[i][0]; beg.b=end.b+move[i][1];//坐标变化即向四个方向移动 if(!vis[beg.a][beg.b]&&0<beg.a&&beg.a<=n&&beg.b>0&&beg.b<=m&&map[beg.a][beg.b]!='S'&&map[beg.a][beg.b]!='R') {//当前点没有走过且坐标不超出题目要求范围 而且此点可以走 vis[beg.a][beg.b]=1; if(map[beg.a][beg.b]=='B') //如果遇到砖墙，花费能量加2 beg.cost=end.cost+2; else beg.cost=end.cost+1; // 否则就是遇见空地 花费加1 q.push(beg); } } } printf("-1 "); } int main() { int j,i,s,t; while(scanf("%d%d",&n,&m)&&n!=0&&m!=0) { for(i=1;i<=n;i++) { getchar(); for(j=1;j<=m;j++) { scanf("%c",&map[i][j]); if(map[i][j]=='Y')//找到起点坐标 { x1=i;y1=j; } else if(map[i][j]=='T')//找到终点坐标 { x2=i;y2=j; } } } memset(vis,0,sizeof(vis)); bfs(); } return 0; }