基本按照刘汝佳书的例题去解.WA了一次：step[front] >10 而不是 step[front]>=10,因为 走10步 可能有多种状态，不应马上return -1;

#include<stdio.h> #include<string.h> const int maxstate=2000000; typedef int State[25]; State que[maxstate]; int step[maxstate]; State goal={1,1,1,1,1,0,1,1,1,1,0,0,2,1,1,0,0,0,0,1,0,0,0,0,0}; State init; const int dx[]={1,2,2,1,-1,-2,-2,-1}; const int dy[]={2,1,-1,-2,2,1,-1,-2}; const int maxhash=1000003; int head[maxhash]; int next[maxstate]; int hash(State& s) { int v=0; for(int i=0;i < 25;i++) { v=v*2+s[i]; } return v%maxhash; } bool tryinsert(int queid) { int h=hash(que[queid]); int fir=head[h]; while(fir) { if(0==memcmp(que[queid],que[fir],sizeof(que[queid]))) return false; fir = next[fir]; } next[queid]=head[h]; head[h]=queid; return true; } int bfs() { // memset(head,0,sizeof(head)); int front=0,rear=0; memcpy(que[rear],init,sizeof(State)); step[rear++]=0; while(front<rear) { State& s = que[front]; if(memcmp(&s,&goal,sizeof(s))==0) return step[front]; if(step[front] >10) //> 不是>= return -1; int z; for(z=0; z<25;z++) if(s[z]==2) break; int x,y; x = z/5; y = z%5; int nx,ny,nz; for(int i=0; i<8;i++) { nx=x+dx[i]; ny=y+dy[i]; nz=nx*5+ny; if(nx>=0&&ny>=0&&nx<5&&ny<5) { State& t =que[rear]; memcpy(&t,&s,sizeof(s)); t[nz]=s[z]; t[z]=s[nz]; step[rear]=step[front]+1; if(tryinsert(rear)) rear++; } } front++; } } int main() { // freopen("input.txt","r",stdin); //freopen("out.txt","w",stdout); int numg; scanf("%d",&numg); while(getchar()!='\n') continue; char ch; while(numg--) { for(int i=0;i<5;i++) { for(int j=0;j<5;j++) { ch=getchar(); if(ch==' ') { init[i*5+j]=2; } else init[i*5+j]=ch-'0'; } while(getchar()!='\n') continue; } int res=bfs(); if(res==-1) puts("Unsolvable in less than 11 move(s)."); else printf("Solvable in %d move(s).\n",res); } }