标签：sel cst ons contain void memory 题解 and cin
Time limit1000 ms
Memory limit30000 kB
Input
Output
Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
题意：红色和黑色的地砖，只能走黑色的地砖，问最多可以走几块地砖
题解：dfs
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <map> #include <set> #include <stack> #include <vector> #include <list> using namespace std; #define PI 3.14159265358979323846264338327950 #define INF 0x3f3f3f3f3f3f3f3f; char a[25][25]; int vis[25][25]; int m,n,st,en,sum; void dfs(int x,int y) { a[x][y]=‘#‘; sum++; if(x-1>=0 && a[x-1][y]==‘.‘) dfs(x-1,y); if(x+1<n && a[x+1][y]==‘.‘) dfs(x+1,y); if(y-1>=0 && a[x][y-1]==‘.‘) dfs(x,y-1); if(y+1<m && a[x][y+1]==‘.‘) dfs(x,y+1); } int main() { while(scanf("%d %d",&m,&n) && (m||n)) { sum=1; int i,j; memset(vis,0,sizeof(vis)); for( i=0;i<n;i++) for(j=0;j<m;j++) { cin>>a[i][j]; if(a[i][j]==‘@‘) { st=i; en=j; } } int x=st,y=en; a[x][y]=‘#‘; if(x-1>=0 && a[x-1][y]==‘.‘) dfs(x-1,y); if(x+1<n && a[x+1][y]==‘.‘) dfs(x+1,y); if(y-1>=0 && a[x][y-1]==‘.‘) dfs(x,y-1); if(y+1<m && a[x][y+1]==‘.‘) dfs(x,y+1); printf("%d\n",sum); } }
标签：sel cst ons contain void memory 题解 and cin
原文地址：https://www.cnblogs.com/smallhester/p/9499285.html
