CodeForces 6B President's Office
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题意:在一个n行m列的办公室中有一些小的矩形桌子,每个人的桌子都有唯一的颜色与之对应,总统的副手与总统桌子相邻,给出一个图,c代表总统桌子颜色,求总统有多少个副手。
思路:从总统的桌子开始找,相邻的桌子,并将其去掉,没去掉一次人数加一次,写一个去掉桌子的函数。
#include<iostream>using namespace std;const int maxx = 101;char map[maxx][maxx];char c;int num,n,m;int cal[5] = {1,0,-1,0,1};void clear(int x,int y);void dfs(int x,int y){if(map[x][y]==c){for(int i=0;i<4;i++)// x+1,y x,y+1 x-1,y x,y-1{int x1=x+cal[i],y1=y-cal[i+1];if(map[x1][y1]!='.'&&x1>=0&&x1<n&&y1>=0&&y1<m){map[x][y]='.';dfs(x1,y1);}}}else if(map[x][y]!='.'){clear(x,y);num++;}}void clear(int x,int y){char b = map[x][y];map[x][y]='.';for(int i=0;i<4;i++){int x1=x+cal[i],y1=y-cal[i+1];if(map[x1][y1]==b&&x1>=0&&x1<n&&y1>=0&&y1<m){clear(x1,y1);}}}int main(){int x,y;while(cin>>n>>m){cin>>c;num=0;for(int i=0;i<n;i++)for(int j=0;j<m;j++){cin>>map[i][j];}for(int i=0;i<n;i++)for(int j=0;j<m;j++){if(map[i][j]==c){x=i;y=j;break;}}dfs(x,y);cout<<num<<endl;}}
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