题目表述: http://acm.hdu.edu.cn/showproblem.php?pid=1253
单纯的多了两个方向的 bfs 而已, 纯当找手感.(然而也改了好几次...
- #include<cstring>
- #include<cstdio>
- using namespace std;
- struct node{
- int a,b,c;
- }q[50005];
- int K,mp[55][55][55],head,tail,A,B,C,T,f[55][55][55],flag;
- int da[6]={1,-1,0,0,0,0};
- int db[6]={0,0,1,-1,0,0};
- int dc[6]={0,0,0,0,1,-1};
- int main(){
- scanf("%d",&K);
- while(K--){
- memset(mp,0,sizeof mp);
- memset(f,0,sizeof f);
- flag = 0;
- int a,b,c,o;
- scanf("%d%d%d%d",&A,&B,&C,&T);
- for(int i=0;i<A;++i){
- for(int j=0;j<B;++j){
- for(int k=0;k<C;++k){
- scanf("%d",&o);
- mp[i][j][k] = o;
- }
- }
- }
- if(mp[A-1][B-1][C-1]){
- printf("-1\n");
- continue;
- }
- q[1].a = q[1].b = q[1].c = 0;
- head = tail = 1;
- f[0][0][0] = 1;
- while(head <= tail){
- a = q[head].a; b = q[head].b; c = q[head].c;
- for(int i=0;i<6;++i){
- int na,nb,nc;
- na = a + da[i]; nb = b + db[i]; nc = c + dc[i];
- if(na>=0 && na<A && nb>=0 && nb<B && nc>=0 && nc<C){
- if(!f[na][nb][nc] && !mp[na][nb][nc] && A+B+C-na-nb-nc-3<=T){
- f[na][nb][nc] = f[a][b][c] + 1;
- if(f[na][nb][nc]>=T) flag = 1;
- q[++tail].a = na; q[tail].b = nb; q[tail].c = nc;
- }
- }
- }
- head++;
- if(f[A-1][B-1][C-1] || flag) break;
- }
- if(!f[A-1][B-1][C-1]||f[A-1][B-1][C-1]>=T){
- printf("-1\n");
- }
- else printf("%d\n",f[A-1][B-1][C-1]-1);
- }
- return 0;
- }
来源: http://www.bubuko.com/infodetail-3211982.html