HDU-1459.非常可乐(BFS )

　　这道题TLE了很多次，原来一直以为将数字化为最简可以让运算更快，但是去了简化之后才发现，真正耗时的就是化简....还和队友学到了用状态少直接数组模拟刚就能过...

　　本题大意：给出可乐的体积v1，给出两个杯子v2和v3，要求v2 + v3 == v1，每次只能从一个杯子倒到另一个杯子，问最少倒多少次可以将可乐平分。

　　思路：最后可乐肯定在可乐瓶和大杯子里面，直接BFS暴搜就行了。

　　参考代码：

 #include <cstdio>
 #include <queue>
 #include <cstring>
 using namespace std;
 
 struct node {
     int cur[], step;
 } now, Next;
 
 const int maxn =  + ;
 int v[];
 int  ans, t;
 bool flag;
 bool vis[maxn][maxn][maxn];
 
 void bfs() {
     if(v[] & ) {
         flag = false;
         return;
     }
     queue <node> Q;
     now.cur[] = v[], now.cur[] = , now.cur[] = , now.step = ;
     vis[v[]][][] = true;
     Q.push(now);
     while(!Q.empty()) {
         now = Q.front();
         Q.pop();
         if(now.cur[] == now.cur[] && now.cur[] == ) {
             ans = now.step;
             return;
         }
         for(int i = ; i < ; i ++) {//i -> j
             for(int j = ; j < ; j ++) {
                 if(i != j) {
                     Next = now;
                     t = now.cur[i] + now.cur[j];
                     if(t > v[j])  Next.cur[j] = v[j];
                     else Next.cur[j] = t;
                     Next.cur[i] = t - Next.cur[j];
                     if(!vis[Next.cur[]][Next.cur[]][Next.cur[]]) {
                         Next.step = now.step + ;
                         Q.push(Next);
                         vis[Next.cur[]][Next.cur[]][Next.cur[]] = true;
                     }
                 }
             }
         }
     }
     flag = false;
 }
 
 int main () {
     while(~scanf("%d %d %d", &v[], &v[], &v[])) {
         memset(vis, false, sizeof(vis));
         if(v[] ==  && v[] ==  && now.cur[] == )  break;
         flag = true, ans = ;
         if(v[] < v[]) {
             t = v[];
             v[] = v[];
             v[] = t;
         }
         bfs();
         if(flag) printf("%d\n", ans);
         else    printf("NO\n");
     }
     return ;
 }

　　下面这个代码的效率要高一点，区别在于倒水方式不同。

 #include <cstdio>
 #include <queue>
 #include <cstring>
 using namespace std;
 
 struct node {
     int cur[], step;
 } now, Next;
 
 const int maxn =  + ;
 int v[];
 int  ans, t;
 bool flag;
 bool vis[maxn][maxn][maxn];
 
 void bfs() {
     if(v[] & ) {
         flag = false;
         return;
     }
     queue <node> Q;
     now.cur[] = v[], now.cur[] = , now.cur[] = , now.step = ;
     vis[v[]][][] = true;
     Q.push(now);
     while(!Q.empty()) {
         now = Q.front();
         Q.pop();
         if(now.cur[] == now.cur[] && now.cur[] == ) {
             ans = now.step;
             return;
         }
         for(int i = ; i < ; i ++) {//i -> j
             for(int j = ; j < ; j ++) {
                 if(i != j) {
                     Next = now;
                     if(j == ) {
                         Next.cur[j] = now.cur[j] + now.cur[i];
                         Next.cur[i] = ;
                     } else if(j == ) {
                         t = v[] - now.cur[j];
                         Next.cur[i] = now.cur[i] - t;
                         Next.cur[i] = Next.cur[i] >  ? Next.cur[i] : ;
                         Next.cur[j] = now.cur[j] + now.cur[i];
                         Next.cur[j] = Next.cur[j] < v[] ? Next.cur[j] : v[];
                     } else {
                         t = v[] - now.cur[j];
                         Next.cur[i] = now.cur[i] - t;
                         Next.cur[i] = Next.cur[i] >  ? Next.cur[i] : ;
                         Next.cur[j] = now.cur[j] + now.cur[i];
                         Next.cur[j] = Next.cur[j] < v[] ? Next.cur[j] : v[];
                     }
                     if(!vis[Next.cur[]][Next.cur[]][Next.cur[]]) {
                         Next.step = now.step + ;
                         Q.push(Next);
                         vis[Next.cur[]][Next.cur[]][Next.cur[]] = true;
                     }
                 }
             }
         }
     }
     flag = false;
 }
 
 int main () {
     while(~scanf("%d %d %d", &v[], &v[], &v[])) {
         memset(vis, false, sizeof(vis));
         if(v[] ==  && v[] ==  && now.cur[] == )  break;
         flag = true, ans = ;
         if(v[] < v[]) {
             t = v[];
             v[] = v[];
             v[] = t;
         }
         bfs();
         if(flag) printf("%d\n", ans);
         else    printf("NO\n");
     }
     return ;
 }