B. Labyrinth
You are playing some computer game. One of its levels puts you in a maze consisting of n lines, each of which contains m cells. Each cell either is free or is occupied by an obstacle. The starting cell is in the row r and column c. In one step you can move one square up, left, down or right, if the target cell is not occupied by an obstacle. You can't move beyond the boundaries of the labyrinth.
Unfortunately, your keyboard is about to break, so you can move left no more than x times and move right no more than y times. There are no restrictions on the number of moves up and down since the keys used to move up and down are in perfect condition.
Now you would like to determine for each cell whether there exists a sequence of moves that will put you from the starting cell to this particular one. How many cells of the board have this property?
The first line contains two integers n, m (1 ≤ n, m ≤ 2000) — the number of rows and the number columns in the labyrinth respectively.
The second line contains two integers r, c (1 ≤ r ≤ n, 1 ≤ c ≤ m) — index of the row and index of the column that define the starting cell.
The third line contains two integers x, y (0 ≤ x, y ≤ 109) — the maximum allowed number of movements to the left and to the right respectively.
The next n lines describe the labyrinth. Each of them has length of m and consists only of symbols '.' and '*'. The j-th character of the i-th line corresponds to the cell of labyrinth at row i and column j. Symbol '.' denotes the free cell, while symbol '*' denotes the cell with an obstacle.
It is guaranteed, that the starting cell contains no obstacles.
Print exactly one integer — the number of cells in the labyrinth, which are reachable from starting cell, including the starting cell itself.
4 5
3 2
1 2
.....
.***.
...**
*....
10
4 4
2 2
0 1
....
..*.
....
....
7
Cells, reachable in the corresponding example, are marked with '+'.
First example:
+++..
+***.
+++**
*+++.
Second example:
.++.
.+*.
.++.
.++.
题意：给出图，给定左转右转的次数限制，问你从起点最多可到达几个点（包括起点）。
思路：直接bfs（有左右次数的限制的bfs）但是！！！，要先走左右转弯花费次数小的（可惜没想到，太菜了），因此可以用优先队列。
剩下的看代码吧。
1 #include<iostream> 2 #include<cstdio> 3 #include<string.h> 4 #include<queue> 5 #define LL long long 6 using namespace std; 7 char map[2010][2010]; 8 bool vis[2010][2010]; 9 int dir[][2]={-1,0,1,0,0,-1,0,1};//上下左右 10 LL n,m,r,c,x,y,cnt; 11 struct node{ 12 LL x,y; 13 LL l,r;//左右 14 friend bool operator < (node a,node b){ 15 return a.l+a.r<b.l+b.r;//剩余左右转次数大的先走 16 } 17 }n1,n2; 18 void bfs(){ 19 priority_queue<node>Q; 20 n1.x=r,n1.y=c; 21 n1.l=x,n1.r=y; 22 Q.push(n1); 23 cnt=1; 24 vis[r][c]=true; 25 while(!Q.empty()){ 26 n2=Q.top(); 27 Q.pop(); 28 for(int i=0;i<4;i++){ 29 n1.x=n2.x+dir[i][0],n1.y=n2.y+dir[i][1]; 30 if(n1.x>=1&&n1.y>=1&&n1.x<=n&&n1.y<=m&&map[n1.x][n1.y]=='.'){ 31 if(!vis[n1.x][n1.y]){ 32 if(i==2){//左 33 if(i==2&&n2.l){ 34 n1.l=n2.l-1; 35 n1.r=n2.r;//这别忘写，我就忘了，找半天 36 vis[n1.x][n1.y]=true; 37 cnt++;//加答案 38 Q.push(n1); 39 } 40 } 41 else if(i==3){//右 42 if(i==3&&n2.r){ 43 n1.r=n2.r-1; 44 n1.l=n2.l;// 还有这，唉，写代码要细心 45 vis[n1.x][n1.y]=true; 46 cnt++; 47 Q.push(n1); 48 } 49 }else{ 50 n1.l=n2.l; 51 n1.r=n2.r; 52 vis[n1.x][n1.y]=true; 53 cnt++; 54 Q.push(n1); 55 } 56 } 57 } 58 } 59 } 60 } 61 int main(){ 62 scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&r,&c,&x,&y); 63 for(int i=1;i<=n;i++){ 64 cin>>map[i]+1; 65 } 66 bfs(); 67 printf("%lld ",cnt); 68 return 0; 69 }
有错请指正。