ZOJ 1002 Fire Net
题目大意:有一个4*4的城市,其中一些格子有墙(X表示墙),在剩余的区域放置碉堡。子弹不能穿透墙壁。问最多可以放置几个碉堡,保证它们不会相互误伤。
解法:从左上的顶点开始遍历,如果这个点不是墙,做深度优先搜索,算出这种情况下的最多碉堡数。每做一次DFS,更新一次最大值。
参考代码:
#include<iostream> #include<cstdio> #include<string.h> using namespace std; char city[6][6]; int i,j,n,ans,maxi,visited[6][6]; void DSF(); int find(int x,int y); int main(){ while(cin>>n&&n){ memset(city,‘X‘,sizeof(city)); //this is included in <string.h> memset(visited,0,sizeof(visited)); ans=maxi=0; getchar(); //this is included in <cstdio.h> for(i=1;i<=n;i++){ for(j=1;j<=n;j++) cin>>city[i][j]; getchar(); } DSF(); cout<<ans<<endl; } return 0; } void DSF(){ int r,c; if(ans<maxi) ans=maxi; for(r=1;r<=n;r++){ //intially it is a loop to sweep all the blocks in the city for(c=1;c<=n;c++){ if(!visited[r][c]&&city[r][c]!=‘X‘){ if(find(r,c)){ // check if can put a castle in this block visited[r][c]=1; //put a castle and the number of castle increase by one maxi++; DSF(); //continue search, similar to pushing this block into stack visited[r][c]=0; //return and number of castle minus one maxi--; } } } } } int find(int x,int y){ int i; for(i=x-1;i>0;i++){ //check if there is castle in front, from the nearest to furthest if(visited[i][y]==1) return 0; if(city[i][y]==‘X‘) break; //if there has a wall, no problem, continue to next direction } for(i=y-1;i>0;i++){ //left if(visited[x][i]==1) return 0; if(city[x][i]==‘X‘) break; } for(i=x+1;i<=n;i++){ //back if(visited[i][y]==1) return 0; if(city[i][y]==‘X‘) break; } for(i=y+1;i<=n;i++){ //right if(visited[x][i]==1) return 0; if(city[x][i]==‘X‘) break; } return 1; }
郑重声明:本站内容如果来自互联网及其他传播媒体,其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享,并不代表本站赞同其观点和对其真实性负责,也不构成任何其他建议。