标签：ble targe accept mem cte std elf square get
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 50913 | Accepted: 27001 |
Description
Input
Output
Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
题意：@是搜索起点，#不能走，.可以走，.走过一次后会变为#，问从@开始在棋盘上一共可以走几步（@起点算一步）
#include<iostream> #include<string.h> #include<string> #include<algorithm> using namespace std; int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}}; string a[105]; int n,m,cnt; int check(int x,int y) { if(x>=0&&x<n&&y>=0&&y<m&&a[x][y]!=‘#‘) return 1; else return 0; } void dfs(int x,int y) { if(check(x,y)==0) return ; else { a[x][y]=‘#‘; cnt++; for(int i=0;i<4;i++) { int dx,dy; dx=x+dir[i][0]; dy=y+dir[i][1]; dfs(dx,dy); } } } int main() { while(cin>>m>>n&&n&&m) { for(int i=0;i<n;i++) cin>>a[i]; cnt=0; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { if(a[i][j]==‘@‘) { dfs(i,j); break; } } } cout<<cnt<<endl; } return 0; }
POJ 1979 Red and Black 四方向棋盘搜索
标签：ble targe accept mem cte std elf square get
原文地址：https://www.cnblogs.com/-citywall123/p/11290235.html
