hdu 1495Legal or Not（搜索+模拟）

刚开始我以为这是一道类似于博弈的题目，还在那儿思考有没有什么办法直接推出答案，看了别人的代码才知道这题没啥好办法，老老实实的模拟才是王道。

应该说是很有意思的搜索，每次根据之前的三个杯子的不同状态产生下一个状态，一直到搜索到结果或者返回-1为止。

#include<stdio.h>#include<string.h>#include<queue>using namespace std;#define N 101int map[N][N][N];int x,y,z;struct node{int x,y,z;int step;};int BFS(){queue<node>q;node cur,next;cur.x=x;cur.y=0;cur.z=0;cur.step=0;q.push(cur);memset(map,0,sizeof(map));map[x][0][0]=1;while(!q.empty()){cur=q.front();q.pop();next.step=cur.step+1;if(cur.x==0&&cur.y==cur.z||cur.y==0&&cur.x==cur.z||cur.z==0&&cur.x==cur.y)return cur.step;if(cur.x){if(cur.x>y-cur.y){next.y=y;next.z=cur.z;next.x=cur.x-(y-cur.y);if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}}else{next.y=cur.y+cur.x;next.x=0;next.z=cur.z;if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}}if(cur.x>z-cur.z){next.z=z;next.x=cur.x-(z-cur.z);next.y=cur.y;if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}}else{next.x=0;next.z=cur.z+cur.x;next.y=cur.y;if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}}}if(cur.y){next.x=cur.x+cur.y;next.y=0;next.z=cur.z;if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}if(cur.y>z-cur.z){next.z=z;next.y=cur.y-(z-cur.z);next.x=cur.x;if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}}else{next.y=0;next.z=cur.z+cur.y;next.x=cur.x;if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}}}if(cur.z){next.x=cur.x+cur.z;next.z=0;next.y=cur.y;if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}if(cur.z>y-cur.y){next.y=y;next.z=cur.z-(y-cur.y);next.x=cur.x;if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}}else{next.y=cur.y+cur.z;next.z=0;next.x=cur.x;if(!map[next.x][next.y][next.z]){q.push(next);map[next.x][next.y][next.z]=1;}}}}return -1;}int main(){int ans;while(scanf("%d%d%d",&x,&y,&z),x||y||z){if(x%2!=0){printf("NO\n");continue;}ans=BFS();if(ans==-1)printf("NO\n");else printf("%d\n",ans);}return 0;}