The Clocks

  这道题应该算是比较经典的一道Brute force method（暴力法），至少我是碰到好多次了。

思路：每一个钟都只有4种状态(3，6，9，12)--->（0，1，2，3),所以每一种操作的数目都会少于4

因此我们需要枚举所有的操作的组合情况就可以了，时间复杂度（4^9)

开始时用一个x【】数组记录每一种操作的步数，对于每一个时钟只有其中的一部分操作对其有影响，所有有影响的加起来应该等于3，这样做效率是高一些，不过太容易出错了，后来参考了别人的代码，重写了一下（注释部分），其实原理还是一样的。建议大家用第二种。

#include<cstdio>#include<iostream>#include<string.h>using namespace std;int x[10],org[10];int ans[10],step;//max=9*3 step=sum(ans[i])bool success(){if( (x[0]+x[1]+x[3]+org[0])%4==3&&//1 (x[0]+x[1]+x[2]+x[4]+org[1])%4==3&&//2(x[1]+x[2]+x[5]+org[2])%4==3&&//3(x[0]+x[3]+x[4]+x[6]+org[3])%4==3&&//4(x[0]+x[2]+x[4]+x[6]+x[8]+org[4])%4==3&&//5(x[2]+x[4]+x[5]+x[8]+org[5])%4==3&&//6(x[3]+x[6]+x[7]+org[6])%4==3&&//7(x[4]+x[6]+x[7]+x[8]+org[7])%4==3&&//8(x[5]+x[7]+x[8]+org[8])%4==3 )return true;//9else return false;}void handle(int cur){if( cur<step )memcpy(ans,x,sizeof(x));else{int i;for(i=0;i<9;i++){if(ans[i]>x[i])break;else if(ans[i]<x[i]){memcpy(ans,x,sizeof(x));break;}}}}void dfs(int cur,int k){//当前已走步数if(cur>step)return;if( k==9 ){if( success() ){handle(cur);}}else{int i;for(i=0;i<4;i++){x[k]=i;dfs(cur+i,k+1);}}}int main(){freopen("clocks.in","r",stdin);freopen("clocks.out","w",stdout);int i;for(i=0;i<9;i++){cin>>org[i];org[i]=org[i]/3-1;}memset(x,0,sizeof(x));memset(ans,0,sizeof(ans));step=27;dfs(0,0);int j,tmp[30],cnt=0;for(i=0;i<9;i++){for(j=0;j<ans[i];j++)tmp[cnt++]=i+1;}for(i=0;i<cnt;i++){cout<<tmp[i];if(i!=cnt-1)cout<<' ';}cout<<endl;return 0;}/*#include<cstdio>#include<iostream>#include<string.h>using namespace std;int flag[9][9]={ //{A,B,C,D,E,F,G,H,I}{1,1,0,1,1,0,0,0,0},{1,1,1,0,0,0,0,0,0},{0,1,1,0,1,1,0,0,0},{1,0,0,1,0,0,1,0,0},{0,1,0,1,1,1,0,1,0},{0,0,1,0,0,1,0,0,1},{0,0,0,1,1,0,1,1,0},{0,0,0,0,0,0,1,1,1},{0,0,0,0,1,1,0,1,1}};int x[10],org[10];int ans[10],step;//max=9*3 step=sum(ans[i])int num[10];void handle(int cur){if( cur<step )memcpy(ans,x,sizeof(x));else{int i;for(i=0;i<9;i++){if(ans[i]>x[i])break;else if(ans[i]<x[i]){memcpy(ans,x,sizeof(x));break;}}}}void dfs(int cur,int k){//当前已走步数if(cur>step)return;if( k==9 ){memset(num,0,sizeof(num));int i,j;for(i=0;i<9;i++){for(j=0;j<9;j++){num[j]+=( x[i]*flag[i][j] ) ;}}bool f=true;for(i=0 ; i<9 && f ; i++)if( (num[i]+org[i])% 4 != 3 )f=false;if( f )handle(cur);}else{int i;for(i=0;i<4;i++){x[k]=i;dfs(cur+i,k+1);}}}int main(){freopen("clocks.in","r",stdin);freopen("clocks.out","w",stdout);int i;for(i=0;i<9;i++){cin>>org[i];org[i]=org[i]/3-1;}step=27;dfs(0,0);int j,tmp[30],cnt=0;for(i=0;i<9;i++){for(j=0;j<ans[i];j++)tmp[cnt++]=i+1;}for(i=0;i<cnt;i++){cout<<tmp[i];if(i!=cnt-1)cout<<' ';}cout<<endl;return 0;}*//* #p1 Test 1: TEST OK [0.022 secs, 2892 KB] Test 2: TEST OK [0.011 secs, 2892 KB] Test 3: TEST OK [0.011 secs, 2892 KB] Test 4: TEST OK [0.011 secs, 2892 KB] Test 5: TEST OK [0.022 secs, 2892 KB] Test 6: TEST OK [0.011 secs, 2892 KB] Test 7: TEST OK [0.011 secs, 2892 KB] Test 8: TEST OK [0.011 secs, 2892 KB] Test 9: TEST OK [0.022 secs, 2892 KB] #p2 Test 1: TEST OK [0.076 secs, 2900 KB] Test 2: TEST OK [0.076 secs, 2900 KB] Test 3: TEST OK [0.076 secs, 2900 KB] Test 4: TEST OK [0.086 secs, 2900 KB] Test 5: TEST OK [0.086 secs, 2900 KB] Test 6: TEST OK [0.097 secs, 2900 KB] Test 7: TEST OK [0.076 secs, 2900 KB] Test 8: TEST OK [0.076 secs, 2900 KB] Test 9: TEST OK [0.076 secs, 2900 KB] */