Hiking in the mountains is seldom an easy task for most people, as it is extremely easy to get lost during the trip. Recently Green has decided to go on a hiking trip. Unfortunately, half way through the trip, he gets extremely tired and so needs to find the path that will bring him to the destination with the least amount of time. Can you help him? You've obtained the area Green's in as an R * C map. Each grid in the map can be one of the four types: tree, sand, path, and stone. All grids not containing stone are passable, and each time, when Green enters a grid of type X (where X can be tree, sand or path), he will spend time T(X). Furthermore, each time Green can only move up, down, left, or right, provided that the adjacent grid in that direction exists. Given Green's current position and his destination, please determine the best path for him.
Input
There are multiple test cases in the input file. Each test case starts with two integers R, C (2 <= R <= 20, 2 <= C <= 20), the number of rows / columns describing the area. The next line contains three integers, VP, VS, VT (1 <= VP <= 100, 1 <= VS <= 100, 1 <= VT <= 100), denoting the amount of time it requires to walk through the three types of area (path, sand, or tree). The following R lines describe the area. Each of the R lines contains exactly C characters, each character being one of the following: ‘T’, ‘.’, ‘#’, ‘@’, corresponding to grids of type tree, sand, path and stone. The final line contains four integers, SR, SC, TR, TC, (0 <= SR < R, 0 <= SC < C, 0 <= TR < R, 0 <= TC < C), representing your current position and your destination. It is guaranteed that Green's current position is reachable – that is to say, it won't be a '@' square. There is a blank line after each test case. Input ends with End-of-File.
Output
For each test case, output one integer on one separate line, representing the minimum amount of time needed to complete the trip. If there is no way for Green to reach the destination, output -1 instead.
Sample Input
4 6
1 2 10
T...TT
TTT###
TT.@#T
..###@
0 1 3 0
4 6
1 2 2
T...TT
TTT###
TT.@#T
..###@
0 1 3 0
2 2
5 1 3
T@
@.
0 0 1 1
Sample Output
Case 1: 14
Case 2: 8
Case 3: -1
题意：给出地图，有path, sand, tree, stone。分别给出经过path, sand, or tree所需的时间，stone是不能通过的。给出起点和终点。问最少的时间达到。不能到达
输出-1,否则输出最少时间。
解析：很明显的优先队列搜索。
代码
#include<cstdio> #include<cstring> #include<string> #include<iostream> #include<sstream> #include<algorithm> #include<utility> #include<vector> #include<set> #include<map> #include<queue> #include<cmath> #include<iterator> #include<stack> using namespace std; const int INF=1e9+7; const double eps=1e-7; const int maxn=25; int row,col,bex,bey,endx,endy; int P,S,T; int dx[]={-1,0,1,0},dy[]={0,-1,0,1}; //方向数组 char maze[maxn][maxn]; //地图 bool vis[maxn][maxn]; //标记数组 bool in(int x,int y){ return x>=0&&x<row&y>=0&&y<col; }//是否在界内 struct node { int x,y,d; node(int x=0,int y=0,int d=0):x(x),y(y),d(d){} bool operator < (const node& t) const { return d>t.d; //越小的优先级越高 } }; priority_queue<node> que; int solve() { memset(vis,false,sizeof(vis)); while(!que.empty()) que.pop(); que.push(node(bex,bey,0)); vis[bex][bey]=true; while(!que.empty()) { node t=que.top(); que.pop(); int x=t.x,y=t.y,d=t.d; if(x==endx&&y==endy) return d; for(int i=0;i<4;i++) { int nx=x+dx[i]; int ny=y+dy[i]; if(!in(nx,ny)||maze[nx][ny]=='@'||vis[nx][ny]) continue; //越界或是石头或已经被标记过 vis[nx][ny]=true; //标记 int nd=d; if(maze[nx][ny]=='T') nd+=T; else if(maze[nx][ny]=='.') nd+=S; else if(maze[nx][ny]=='#') nd+=P; que.push(node(nx,ny,nd)); //丢到优先队列里去 } } return -1; } int main() { int Case=0; while(scanf("%d%d",&row,&col)!=EOF) { scanf("%d%d%d",&P,&S,&T); for(int i=0;i<row;i++) scanf("%s",maze[i]); scanf("%d%d%d%d",&bex,&bey,&endx,&endy); int ans=solve(); printf("Case %d: %d ",++Case,ans); } return 0; }