hdu1760A New Tetris Game-----找sg字符串A New Tetris GameTime Limit: 3000/1000 MS (Java/Others)Memor
hdu1760 A New Tetris Game-----找sg 字符串
A New Tetris Game
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 789 Accepted Submission(s): 343
Problem DescriptionInputOutputSample InputSample OutputAuthorSourceRecommend#include<iostream>#include<cstdlib>#include<stdio.h>#include<string.h>#include<string>#include<map>#include<memory.h>using namespace std;map<string,int>sg;map<string,bool>h;char str[51][51],s[2505];int n,m;bool judge(int x,int y){ if(str[x][y+1]=='0'&&str[x+1][y]=='0'&&str[x+1][y+1]=='0') return true; return false;}void change(int x,int y,char c){ str[x][y]=c;str[x+1][y]=c;str[x][y+1]=c;str[x+1][y+1]=c;}void zhuanhuan(){ strcpy(s,""); for(int i=0;i<n;i++) strcat(s,str[i]);}int search(){ int num[50]; memset(num,0,sizeof(num)); zhuanhuan();h[s]=1; for(int i=0;i<n-1;i++) for(int j=0;j<m-1;j++) { if(str[i][j]=='0') { if(judge(i,j)) { change(i,j,'1');zhuanhuan(); int k; if(h[s]) k=sg[s]; else k=search(); num[k]=1; change(i,j,'0'); } } } for(int i=0;;i++) if(num[i]==0) {zhuanhuan();sg[s]=i;return i; }}int main(){ while(scanf("%d%d",&n,&m)!=EOF) { for(int i=0;i<n;i++) scanf("%s",str[i]); sg.clear();h.clear(); int ans=search(); if(ans) puts("Yes"); else puts("No"); } return 0;}/*4 400000000000000004 40000001001000000*/
法二、利用sg的性质#include<iostream>#include<cstdlib>#include<stdio.h>#include<string.h>#include<string>#include<map>#include<memory.h>using namespace std;char str[51][51];int n,m;bool judge(int x,int y){ if(str[x][y+1]=='0'&&str[x+1][y]=='0'&&str[x+1][y+1]=='0') return true; return false;}void change(int x,int y,char c){ str[x][y]=c;str[x+1][y]=c;str[x][y+1]=c;str[x+1][y+1]=c;}int search(){ int flag=0; for(int i=0;i<n-1;i++) for(int j=0;j<m-1;j++) { if(str[i][j]=='0') { if(judge(i,j)) { flag++; change(i,j,'1'); if(!search()) { change(i,j,'0'); return 1; } change(i,j,'0'); } } } if(flag<=1) return flag; return 0;}int main(){ while(scanf("%d%d",&n,&m)!=EOF) { for(int i=0;i<n;i++) scanf("%s",str[i]); int ans=search(); if(ans) puts("Yes"); else puts("No"); } return 0;}/*4 400000000000000004 40000001001000000*/