九度 OJ 1457 非常可乐

#include <stdio.h>
#include <queue>
using namespace std;

struct P{
int x , y , z;
int cost;
};

queue<P> Q;
bool mark[101][101][101];

void AtoB(int &a , int sa , int &b , int sb)//倾倒函数
//sa,sb为杯子体积；a,b,初始为杯中体积，调用后为新体积
{
if(sb - b >= a){
b += a;
a = 0;
}
else{
a -= sb - b;
b = sb;
}
}

int BFS(int s , int n , int m)
{
while(!Q.empty()){
P now=Q.front();
Q.pop();
int a , b , c;

a = now.x; // 1.a倒b
b = now.y;
c = now.z;
AtoB(a,s,b,n);
if(mark[a][b][c] == false){
mark[a][b][c] = true;
P newp;
newp.x = a;
newp.y = b;
newp.z = c;
newp.cost = now.cost + 1;
if(a == s / 2 && b == s / 2) return newp.cost;//若当前状态已平分，直接返回耗时
if(a == s / 2 && c == s / 2) return newp.cost;
if(b == s / 2 && c == s / 2) return newp.cost;
Q.push(newp);
}

a = now.x; //2.b倒a---与1类似，下同。
b = now.y;
c = now.z;
AtoB(b,n,a,s);
if(mark[a][b][c] == false){
mark[a][b][c] = true;
P newp;
newp.x = a;
newp.y = b;
newp.z = c;
newp.cost = now.cost + 1;
if(a == s / 2 && b == s / 2) return newp.cost;
if(a == s / 2 && c == s / 2) return newp.cost;
if(b == s / 2 && c == s / 2) return newp.cost;
Q.push(newp);
}

a = now.x; //3.a倒c
b = now.y;
c = now.z;
AtoB(a,s,c,m);
if(mark[a][b][c] == false){
mark[a][b][c] = true;
P newp;
newp.x = a;
newp.y = b;
newp.z = c;
newp.cost = now.cost + 1;
if(a == s / 2 && b == s / 2) return newp.cost;
if(a == s / 2 && c == s / 2) return newp.cost;
if(b == s / 2 && c == s / 2) return newp.cost;
Q.push(newp);
}

a = now.x; //4.c倒a
b = now.y;
c = now.z;
AtoB(c,m,a,s);
if(mark[a][b][c] == false){
mark[a][b][c] = true;
P newp;
newp.x = a;
newp.y = b;
newp.z = c;
newp.cost = now.cost + 1;
if(a == s / 2 && b == s / 2) return newp.cost;
if(a == s / 2 && c == s / 2) return newp.cost;
if(b == s / 2 && c == s / 2) return newp.cost;
Q.push(newp);
}

a = now.x; //5.b倒c
b = now.y;
c = now.z;
AtoB(b,n,c,m);
if(mark[a][b][c] == false){
mark[a][b][c] = true;
P newp;
newp.x = a;
newp.y = b;
newp.z = c;
newp.cost = now.cost + 1;
if(a == s / 2 && b == s / 2) return newp.cost;
if(a == s / 2 && c == s / 2) return newp.cost;
if(b == s / 2 && c == s / 2) return newp.cost;
Q.push(newp);
}

a = now.x; //6.c倒b
b = now.y;
c = now.z;
AtoB(c,m,b,n);
if(mark[a][b][c] == false){
mark[a][b][c] = true;
P newp;
newp.x = a;
newp.y = b;
newp.z = c;
newp.cost = now.cost + 1;
if(a == s / 2 && b == s / 2) return newp.cost;
if(a == s / 2 && c == s / 2) return newp.cost;
if(b == s / 2 && c == s / 2) return newp.cost;
Q.push(newp);
}
}
return -1;
}

int main()
{
int s , n , m;
while(scanf("%d%d%d",&s,&n,&m) != EOF){
if(s == 0) break;
if(s % 2 == 1){ //S为奇一定不可能平分
printf("NO\n");
continue;
}

for(int i = 0 ; i <= s ; ++i){
for(int j = 0 ; j <= n ; ++j){
for(int k = 0 ; k <= m ; ++k){
mark[i][j][k] = false;
}
}
}

P tmp;
tmp.x = s;
tmp.y = 0;
tmp.z = 0;
tmp.cost = 0;
while(!Q.empty()) Q.pop();
mark[s][0][0] = true;
Q.push(tmp);

int ans = BFS(s,n,m);
if(ans > 0) printf("%d\n",ans);
else printf("NO\n");
}
return 0;
}