思路：搜索题，第一次做这种类型的题目吧，一开始表示不怎么明白题意所说的东东。其实就是要你判断可乐能不能被平分........
有六种状态，从a瓶到b瓶，a-->c
b-->a b-->c
c-->a c-->b
然后每种状态里面又分两种不同情况，可以将此瓶的水全部清空，不能清空......
然后广搜就可以了........
#include#include #include #include using namespace std;int vist[105][105][105],a,b,c;struct node{ int a,b,c; int step;}s[105];int sum=0;void bfs(){ queue q; memset(vist,0,sizeof(vist)); node p1; p1.a=a; p1.b=0; p1.c=0; p1.step=0; q.push(p1); vist[p1.a][0][0]=1; while(!q.empty()) { p1=q.front(); q.pop(); if((p1.a==a/2&&p1.b==a/2)||(p1.a==a/2&&p1.c==a/2)||(p1.b==a/2&&p1.c==a/2)) { printf("%d\n",p1.step); return; } node p2; if(p1.a!=0) { if(p1.a>b-p1.b) { p2.a=p1.a-(b-p1.b); p2.b=b; p2.c=p1.c; p2.step=p1.step+1; } else { p2.a=0; p2.b=p1.b+p1.a; p2.c=p1.c; p2.step=p1.step+1; } if(!vist[p2.a][p2.b][p2.c]) { vist[p2.a][p2.b][p2.c]=1; q.push(p2); } } if(p1.a!=0) { if(p1.a>c-p1.c) { p2.a=p1.a-(c-p1.c); p2.b=p1.b; p2.c=c; p2.step=p1.step+1; } else { p2.a=0; p2.b=p1.b; p2.c=p1.c+p1.a; p2.step=p1.step+1; } if(!vist[p2.a][p2.b][p2.c]) { vist[p2.a][p2.b][p2.c]=1; q.push(p2); } } if(p1.b!=0) { if(p1.b>a-p1.a) { p2.b=p1.b-(a-p1.a); p2.a=a; p2.c=p1.c; p2.step=p1.step+1; } else { p2.b=0; p2.a=p1.a+p1.b; p2.c=p1.c; p2.step=p1.step+1; } if(!vist[p2.a][p2.b][p2.c]) { vist[p2.a][p2.b][p2.c]=1; q.push(p2); } } if(p1.b!=0) { if(p1.b>c-p1.c) { p2.b=p1.b-(c-p1.c); p2.a=p1.a; p2.c=c; p2.step=p1.step+1; } else { p2.b=0; p2.a=p1.a; p2.c=p1.c+p1.b; p2.step=p1.step+1; } if(!vist[p2.a][p2.b][p2.c]) { vist[p2.a][p2.b][p2.c]=1; q.push(p2); } } if(p1.c!=0) { if(p1.c>a-p1.a) { p2.c=p1.c-(a-p1.a); p2.a=a; p2.b=p1.b; p2.step=p1.step+1; } else { p2.c=0; p2.a=p1.a+p1.c; p2.b=p1.b; p2.step=p1.step+1; } if(!vist[p2.a][p2.b][p2.c]) { vist[p2.a][p2.b][p2.c]=1; q.push(p2); } } if(p1.c!=0) { if(p1.c>b-p1.b) { p2.c=p1.c-(b-p1.b); p2.a=p1.a; p2.b=b; p2.step=p1.step+1; } else { p2.c=0; p2.a=p1.a; p2.b=p1.b+p1.c; p2.step=p1.step+1; } if(!vist[p2.a][p2.b][p2.c]) { vist[p2.a][p2.b][p2.c]=1; q.push(p2); } } } printf("NO\n");}int main(){ while(scanf("%d%d%d",&a,&b,&c)>0&&(a+b+c)) { if(a%2==1) { printf("NO\n"); continue; } bfs(); } return 0;}