POJ 2965 The Pilots Brothers' refrigerator

这个题是一个枚举题。

用的是八皇后算法。

具体操作不说了。

下面是代码：

#include <stdio.h>
struct node
{int x,y;
}de[200],dr[200];
int a[17],b,w,min,cut;
void ro(int i)
{int j=4;while(i-j>=0){if(a[i-j]==0){a[i-j]=1;w--;b++;}else{a[i-j]=0;w++;b--;}j+=4;}j=4;while(i+j<16){if(a[i+j]==0){a[i+j]=1;w--;b++;}else{a[i+j]=0;w++;b--;}j+=4;}
}
void co(int i)
{int j=1;while(i+j<16&&i/4==(i+j)/4){if(a[i+j]==0){a[i+j]=1;w--;b++;}else{a[i+j]=0;w++;b--;}j++;}j=1;while(i-j>=0&&i/4==(i-j)/4){if(a[i-j]==0){a[i-j]=1;w--;b++;}else{a[i-j]=0;w++;b--;}j++;}
}
void fib(int i)
{int j,g;if(i==-1){for(j=i+1; j<16; j++){fib(j);}}else{if(cut==0&&b==0){min=cut;return ;}if(a[i]==0){a[i]=1;w--;b++;}else{a[i]=0;w++;b--;}ro(i);co(i);de[cut].x=i/4;de[cut].y=i%4;cut++;if(b==0&&min>cut){for(g=0;g<cut;g++){dr[g]=de[g];}min=cut;}else{for(j=i+1; j<16; j++){fib(j);}}if(a[i]==0){a[i]=1;w--;b++;}else{a[i]=0;w++;b--;}ro(i);co(i);cut--;}
}
int main()
{int i,j;char s[6];while(scanf("%s",s)!=EOF){b=0;w=0;min=1<<30;cut=0;for(i=0; i<4; i++){if(s[i]=='+'){a[i]=1;b++;}else if(s[i]=='-'){a[i]=0;w++;}}for(i=1; i<4; i++){scanf("%s",s);for(j=0; j<4; j++){if(s[j]=='+'){a[i*4+j]=1;b++;}else if(s[j]=='-'){a[i*4+j]=0;w++;}}}fib(-1);printf("%d\n",min);for(i=0; i<min; i++){printf("%d %d\n",dr[i].x+1,dr[i].y+1);}}return 0;
}

转载于:https://www.cnblogs.com/lin375691011/p/3996769.html

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