USACO Section 2.2 Party Lamps - 看清本质就简单!

这题乍一看吓一跳...范围挺大的...想了也有一个小时才想通...看清本质!!

总共就4个按钮...如果我1,2,3,4都按过一次了..再按一下1 or 2 or 3 or 4灯的情况不和只按了2,3,4 or 1,3,4 or 1,2,4 or 1,2,3一样吗?因为一个按钮按两次就相当于没按..那么实际上能出现的最多情况也就是C(1,4)+C(2,4)+C(3,4)+C(4,4)=17种~~枚举出这些情况..当且仅当当前所案件的次数与所需要的按键次数之差为偶数(不断地来回按)并且得到的情况是满足输入的要求的就找到了一个~~

输出的话再排个序行了~~

Program:

/* ID: zzyzzy12 LANG: C++ TASK: lamps */ #include<iostream> #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<queue> using namespace std; struct node { int s[102]; }ans[1001],h; int N,C,x,s[102],num; bool f; bool cmp(node a,node b) { int i; for (i=1;i<=N;i++) if (a.s[i]!=b.s[i]) return a.s[i]<b.s[i]; } void search(node h,int step,int key) { int i,j; node k; if (step<=C && (C-step)%2==0) { for (i=1;i<=N;i++) if (s[i]!=-1 && s[i]!=h.s[i]) goto A; ans[++num]=h; if (num!=1) { for (i=1;i<num;i++) { f=true; for (j=1;j<=N;j++) if (ans[num].s[j]!=ans[i].s[j]) f=false; if (f) break; } if (f) num--; } } A: ; for (i=key;i<=4;i++) { if (i==1) { k=h; for (j=1;j<=N;j++) k.s[j]=1-k.s[j]; } else if (i==2) { k=h; for (j=1;j<=N;j++) if (j%2) k.s[j]=1-k.s[j]; } else if (i==3) { k=h; for (j=1;j<=N;j++) if (j%2==0) k.s[j]=1-k.s[j]; } else if (i==4) { k=h; for (j=1;j<=N;j++) if (j%3==1) k.s[j]=1-k.s[j]; } search(k,step+1,key+1); } } int main() { freopen("lamps.in","r",stdin); freopen("lamps.out","w",stdout); scanf("%d%d",&N,&C); memset(s,-1,sizeof(s)); while (scanf("%d",&x) && x!=-1) s[x]=1; while (scanf("%d",&x) && x!=-1) s[x]=0; num=0; for (int i=1;i<=N;i++) h.s[i]=1; search(h,0,1); if (num) { sort(ans+1,ans+1+num,cmp); int j,i; for (i=1;i<=num;i++) { for (j=1;j<N;j++) printf("%d",ans[i].s[j]); printf("%d\n",ans[i].s[N]); } }else printf("IMPOSSIBLE\n"); return 0; }