HDU

非常可乐

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14827    Accepted Submission(s): 5950

Problem 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 34 1 30 0 0

 

Sample Output

NO3

 

Author

seeyou

 

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

 

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LL

你有三个容量分别为S, M, N的无刻度瓶子且保证S == M + N, 初始状态下S装满了可乐，问最少倒多少次能使得S， M， N 中

两个容器各装为S / 2体积的可乐。

直接BFS疯狂赋值黏贴模拟过程就好了，一个容器要倒到另一个容器就两种情况，要不然一个瓶子倒空，要不然另一个瓶子倒满，溢出也好负数也好直接通过一个check剪掉就好了。。就不用一直用if  去 judge了

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<vector>#include<set>#include<map>#include<queue>#include<stack>using namespace std;int const N = 100010;int n, m, w;int arm;int visa[200], visb[200], visc[200];int mark;struct Node{    int a;    int b;    int c;    int ans;};bool judge(Node x){    if(visa[x.a] && visb[x.b] && visc[x.c]) return false;    if(x.a > w || x.b > n || x.c > m) return false;    if(x.a < 0 || x.b < 0 || x.c < 0) return false;    return true;}int bfs(Node x){    queue<Node> q;    Node now;    Node next;    int t;    int ans = 0;    q.push(x);    while(!q.empty()){        now = q.front();        q.pop();        if(now.a == arm && now.b == arm || now.a == arm && now.c == arm || now.b == arm && now.c == arm){            return now.ans;        }        if(judge(now) == false) continue;        visa[now.a] = 1;        visb[now.b] = 1;        visc[now.c] = 1;        t = n - now.b;  //1 - 2        next.a = 0;     //1Пе        next.b = now.b + now.a;        next.c = now.c;        next.ans = now.ans + 1;        q.push(next);                next.a = now.a - t; //2Тњ        next.b = n;        next.c = now.c;        next.ans = now.ans + 1;        q.push(next);        t = m - now.c;      //1 - 3        next.a = 0;     //1Пе        next.c = now.c + now.a;        next.b = now.b;        next.ans = now.ans + 1;        q.push(next);                next.a = now.a - t; //2Тњ        next.c = m;        next.b = now.b;        next.ans = now.ans + 1;        q.push(next);        t = w - now.a;  //2 - 1        next.b = 0;     //2Пе        next.a = now.b + now.a;        next.c = now.c;        next.ans = now.ans + 1;        q.push(next);            next.b = now.b - t; //1Тњ        next.a = w;        next.c = now.c;        next.ans = now.ans + 1;        q.push(next);        t = m - now.c;      //2 - 3        next.b = 0;     //2Пе        next.c = now.c + now.b;        next.a = now.a;        next.ans = now.ans + 1;        q.push(next);        next.b = now.b - t; //3Тњ        next.c = m;        next.a = now.a;        next.ans = now.ans + 1;        q.push(next);        t = w - now.a;  //3 - 1        next.c = 0;     //3Пе        next.a = now.c + now.a;        next.b = now.b;        next.ans = now.ans + 1;        q.push(next);        next.c = now.c - t; //1Тњ        next.a = w;        next.b = now.b;        next.ans = now.ans + 1;        q.push(next);        t = n - now.b;      //3 - 2        next.c = 0;     //3Пе        next.b = now.c + now.b;        next.a = now.a;        next.ans = now.ans + 1;        q.push(next);                next.c = now.c - t; //2Тњ        next.b = n;        next.a = now.a;        next.ans = now.ans + 1;        q.push(next);                ans ++;    }    return 0;}int main(){    int a, b;    Node first;    int ans;    while(scanf("%d%d%d", &w, &n, &m) == 3){        if(w % 2){            printf("NO\n");            continue;        }        if(!n && !m && !w) break;        arm = w / 2;        memset(visa, 0, sizeof(visa));        memset(visb, 0, sizeof(visb));        memset(visc, 0, sizeof(visc));        first.a = w;        first.b = 0;        first.c = 0;        first.ans = 0;        ans = bfs(first);        if(ans)  printf("%d\n", ans);        else printf("NO\n");    }    return 0;}

后来再翻这道题的时候发现有数论的做法，然而我还看不懂。。。代码短的一比。。。。惭愧。。。还有很长的路要走。。。这样的代码才叫做ACM啊

http://m.blog.csdn.net/article/details?id=52097459