題目大意：有一瓶可樂，已知容積，但是不知道刻度，現在有兩個杯子，可以從一個杯子倒到另一個杯子
問最少幾次可以把可樂評分 兩個杯子容積之和爲可樂容積
思路：有大佬推出了公式可以直接推出 次數a/gcd（b，c）-1 如果這個數是偶數則可以輸出，不然輸出no
另一個就是bfs的，bfs時走法很多，比較繁瑣，都是複製粘貼。菜雞寫的很長 還差點超時
#include<stdio.h>
#include<string.h>
int book[105][105][105];
int x,y,z;
struct node{
int a;
int b;
int c;
int step;
}queue[1000005];
int bfs(int i,int j,int k);
int main(void)
{
while(scanf("%d%d%d",&x,&y,&z),x&&y&&z){
if(y<z){
int t;
t=y;
y=z;
z=t;
}
memset(book,0,sizeof(book));
int ans;
if(x%2==0)//判斷是不是偶數，不是的話直接輸出NO
ans=bfs(x,0,0);
else
ans=0;
if(ans)
printf("%d\n",ans);
else
printf("NO\n");
}
return 0;
}
int bfs(int i,int j,int k){
book[i][j][k]=1;
int head=0;
int tail=1;
queue[head].a=i;
queue[head].b=j;
queue[head].c=k;
queue[head].step=0;
while(head<tail){
if(queue[head].a==x/2&&queue[head].b==x/2)//判斷是不是已經平分
return queue[head].step;
if(queue[head].a!=0){
if(queue[head].a>y-queue[head].b) //下面就是pour，看起來很繁瑣，其實都是複製粘貼過去的
{
if(!book[queue[head].a-y+queue[head].b][y][queue[head].c]){
queue[tail].a=queue[head].a-y+queue[head].b;
queue[tail].b=y;
queue[tail].c=queue[head].c;
queue[tail].step=queue[head].step+1;
book[queue[head].a-y+queue[head].b][y][queue[head].c]=1;
if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
}
}
else
{
if(!book[0][queue[head].a+queue[head].b][queue[head].c]){
queue[tail].a=0;
queue[tail].b=queue[head].a+queue[head].b;
queue[tail].c=queue[head].c;
queue[tail].step=queue[head].step+1;if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
book[0][queue[head].a+queue[head].b][queue[head].c]=1;
}
}
if(queue[head].a>z-queue[head].c)
{
if(!book[queue[head].a-z+queue[head].c][queue[head].b][z]){
queue[tail].a=queue[head].a-z+queue[head].c;
queue[tail].c=z;
queue[tail].b=queue[head].b;
queue[tail].step=queue[head].step+1;
book[queue[head].a-z+queue[head].c][queue[head].b][z]=1;if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
}
}
else
{
if(!book[0][queue[head].b][queue[head].a+queue[head].c]){
queue[tail].a=0;
queue[tail].b=queue[head].b;
queue[tail].c=queue[head].a+queue[head].c;
queue[tail].step=queue[head].step+1;if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
book[0][queue[head].b][queue[head].a+queue[head].c]=1;
}
}
}
if(queue[head].b!=0){
if(!book[queue[head].a+queue[head].b][0][queue[head].c]){
queue[tail].a=queue[head].a+queue[head].b;
queue[tail].b=0;
queue[tail].c=queue[head].c;
queue[tail].step=queue[head].step+1;if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
book[queue[head].a+queue[head].b][0][queue[head].c]=1;
}
if(queue[head].b>(z-queue[head].c))
{
if(!book[queue[head].a][queue[head].b-z+queue[head].c][z]){
queue[tail].a=queue[head].a;
queue[tail].c=z;
queue[tail].b=queue[head].b-z+queue[head].c;
queue[tail].step=queue[head].step+1;
book[queue[head].a][queue[head].b-z+queue[head].c][z]=1;if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
}
}
else
{
if(!book[queue[head].a][0][queue[head].b+queue[head].c]){
queue[tail].a=queue[head].a;
queue[tail].b=0;
queue[tail].c=queue[head].b+queue[head].c;
queue[tail].step=queue[head].step+1;if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
book[queue[head].a][0][queue[head].b+queue[head].c]=1;
}
}
}
if(queue[head].c!=0){
if(!book[queue[head].a+queue[head].c][queue[head].b][0]){
queue[tail].a=queue[head].a+queue[head].c;
queue[tail].b=queue[head].b;
queue[tail].c=0;
queue[tail].step=queue[head].step+1;
book[queue[head].a+queue[head].c][queue[head].b][0]=1;if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
}
if(queue[head].c>(y-queue[head].b))
{
if(!book[queue[head].a][y][queue[head].c-y+queue[head].b]){
queue[tail].a=queue[head].a;
queue[tail].c=queue[head].c-y+queue[head].b;
queue[tail].b=y;
queue[tail].step=queue[head].step+1;
book[queue[head].a][y][queue[head].c-y+queue[head].b]=1; if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
}
}
else
{
if(!book[queue[head].a][queue[head].b+queue[head].c][0]){
queue[tail].a=queue[head].a;
queue[tail].b=queue[head].b+queue[head].c;
queue[tail].c=0;
queue[tail].step=queue[head].step+1; if(queue[tail].a==x/2&&queue[tail].b==x/2)
return queue[tail].step;
tail++;
book[queue[head].a][queue[head].b+queue[head].c][0]=1;
}
}
}
head++;
}
return 0;
}