HDU - 1495 十分可乐
HDU - 1495 非常可乐
题意:
题意:
Description
大家一定觉的运动以后喝可乐是一件很惬意的事情,但是seeyou却不这么认为。因为每次当seeyou买了可乐以后,阿牛就要求和seeyou一起分享这一瓶可乐,而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子,它们的容量分别是N 毫升和M 毫升 可乐的体积为S (S<101)毫升 (正好装满一瓶) ,它们三个之间可以相互倒可乐 (都是没有刻度的,且 S==N+M,101>S>0,N>0,M>0) 。聪明的ACMER你们说他们能平分吗?如果能请输出倒可乐的最少的次数,如果不能输出"NO"。
Input
三个整数 : S 可乐的体积 , N 和 M是两个杯子的容量,以"0 0 0"结束。
Output
如果能平分的话请输出最少要倒的次数,否则输出"NO"。
Sample Input
7 4 3 4 1 3 0 0 0
Sample Output
NO 3 思路:BFS的思路,一共六种情况#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; const int MAXN = 110; int s,m,n; int q[MAXN*MAXN*MAXN][3]; int dp[MAXN][MAXN][MAXN]; int bfs(){ if (s & 1) return 0; if (n < (s >> 1)) return 0; dp[s][0][0] = 0; int front = 0,rear = 1; q[front][0] = s,q[front][1] = 0,q[front][2] = 0; while (front < rear){ int x,y,z; int nx,ny,nz; int cur; x = q[front][0],y = q[front][1],z = q[front][2]; front++; // s->n if (x != 0 && y != n){ cur = min(x,n-y); nx = x - cur; ny = y + cur; nz = z; if (dp[nx][ny][nz] == -1){ dp[nx][ny][nz] = dp[x][y][z] + 1; q[rear][0] = nx,q[rear][1] = ny,q[rear][2] = nz; rear++; } } // s->m if (x != 0 && z != m){ cur = min(x,m-z); nx = x - cur; ny = y; nz = z + cur; if (dp[nx][ny][nz] == -1){ dp[nx][ny][nz] = dp[x][y][z] + 1; q[rear][0] = nx,q[rear][1] = ny,q[rear][2] = nz; rear++; } } // n->s if (y != 0 && x != s){ cur = min(y,s-x); nx = x + cur; ny = y - cur; nz = z; if (dp[nx][ny][nz] == -1){ dp[nx][ny][nz] = dp[x][y][z] + 1; q[rear][0] = nx,q[rear][1] = ny,q[rear][2] = nz; rear++; } } // n->m if (y != 0 && z != m){ cur = min(y,m-z); nx = x; ny = y - cur; nz = z + cur; if (dp[nx][ny][nz] == -1){ dp[nx][ny][nz] = dp[x][y][z] + 1; q[rear][0] = nx,q[rear][1] = ny,q[rear][2] = nz; rear++; } } // m->s if (z != 0 && x != s){ cur = min(z,s-x); nx = x + cur; ny = y; nz = z - cur; if (dp[nx][ny][nz] == -1){ dp[nx][ny][nz] = dp[x][y][z] + 1; q[rear][0] = nx,q[rear][1] = ny,q[rear][2] = nz; rear++; } } // m->n if (z != 0 && y != n){ cur = min(z,n-y); nx = x; ny = y + cur; nz = z - cur; if (dp[nx][ny][nz] == -1){ dp[nx][ny][nz] = dp[x][y][z] + 1; q[rear][0] = nx,q[rear][1] = ny,q[rear][2] = nz; rear++; } } } return dp[s>>1][s>>1][0]; } int main(){ while (scanf("%d%d%d",&s,&n,&m) != EOF && s+n+m){ if (n < m) swap(n,m); memset(dp,-1,sizeof(dp)); int ans = bfs(); if (ans <= 0) printf("NO\n"); else printf("%d\n",ans); } return 0; }