HDU - 1495 bfs

这篇文章用的bfs，昨天还看到一个大神的数论推导，太强了：https://blog.csdn.net/v5zsq/article/details/52097459

这一道题由于没有给出杯子的具体量度，只是给出了最大容量，故每次两个杯子之间的状态为两种：

       1. s1全部倒入s2中；(s1现在的液体和s2中已有的液体之和小于等于s2的容积)

       2.s1倒入s2中，s1中仍有剩余；(s1已有的液体与s2中已有的液体之和超过了s2的容积，故s1有剩余)

只有三个容器，因此如果满足条件（假设a->b,需要a中有液体，而b未满）即可互相倾倒，因此某一步最多可以转移出六种状态：即a->b,a->s,b->a,b->s,s->a,s->b;

那既然需要计算最小步数就需要一个step数组记录状态，对于未出现的状态step[this.a][this.a]=step[last.a][last.b]+1;

show u code：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const double epos=1e-8;
const int maxn=109;

//记录状态出现与否;因为题目明确指出s==a+b,故两个维度表示a,b第三个维度s一定是确定的;
int flag[maxn][maxn];

//记录步数,对于未出现的状态step[this.a][this.a]=step[last.a][last.b]+1;
int step[maxn][maxn];
int s,a,b;//将a看做大杯子,当然也可以不管只不过是bfs退出条件那里多了一条语句而已;

struct Node{
    int a,b,s;
};//分别表示当前杯子中的量;
queue<Node>q;

void init(){
    memset(flag,0,sizeof(flag));
    memset(step,0,sizeof(step));
    while(!q.empty())
        q.pop();
}

int bfs(){
    init();
    Node chushi,now,in;//now是每次从队列中弹出的状态,in是要进入队列的状态;
    chushi.a=0,chushi.b=0,chushi.s=s;
    flag[chushi.a][chushi.b]=1;
    q.push(chushi);
    while(!q.empty()){
        now=q.front();
        q.pop();
        if(now.a==s/2&&now.s==s/2)
            return step[now.a][now.b];
        if(now.a&&now.b<b){
            if(now.a+now.b<b){
                in.a=0;
                in.s=now.s;
                in.b=now.a+now.b;
            }
            else{
                in.b=b;
                in.s=now.s;
                in.a=now.a-b+now.b;
            }
            if(flag[in.a][in.b]==0){
                step[in.a][in.b]=step[now.a][now.b]+1;
                q.push(in);
                flag[in.a][in.b]=1;
            }
        }//a->b


        if(now.a&&now.s<s){
            if(now.a+now.s<s){
                in.a=0;
                in.s=now.a+now.s;
                in.b=now.b;
            }
            else{
                in.s=s;
                in.a=now.a-s+now.s;
                in.b=now.b;
            }
            if(flag[in.a][in.b]==0){
                step[in.a][in.b]=step[now.a][now.b]+1;
                q.push(in);
                flag[in.a][in.b]=1;
            }
        }//a->s


        if(now.b&&now.a<a){
            if(now.a+now.b<a){
                in.b=0;
                in.a=now.a+now.b;
                in.s=now.s;
            }
            else{
                in.a=a;
                in.b=now.b-a+now.a;
                in.s=now.s;
            }

            if(flag[in.a][in.b]==0){
                step[in.a][in.b]=step[now.a][now.b]+1;
                q.push(in);
                flag[in.a][in.b]=1;
            }
        }//b->a


        if(now.b&&now.s<s){
            if(now.b+now.s<s){
                in.a=now.a;
                in.b=0;
                in.s=now.b+now.s;
            }
            else{
                in.a=now.a;
                in.b=now.b-s+now.s;
                in.s=s;
            }
            if(flag[in.a][in.b]==0){
                step[in.a][in.b]=step[now.a][now.b]+1;
                q.push(in);
                flag[in.a][in.b]=1;
            }
        }//b->s


        if(now.s&&now.a<a){
            if(now.s+now.a<a){
                in.s=0;
                in.b=now.b;
                in.a=now.a+now.s;
            }
            else{
                in.b=now.b;
                in.a=a;
                in.s=now.s-a+now.a;
            }
            if(flag[in.a][in.b]==0){
                step[in.a][in.b]=step[now.a][now.b]+1;
                q.push(in);
                flag[in.a][in.b]=1;
            }
        }//s->a


        if(now.s&&now.b<b){
            if(now.s+now.b<b){
                in.a=now.a;
                in.b=now.s+now.b;
                in.s=0;
            }
            else{
                in.a=now.a;
                in.b=b;
                in.s=now.s-b+now.b;
            }
            if(flag[in.a][in.b]==0){
                step[in.a][in.b]=step[now.a][now.b]+1;
                q.push(in);
                flag[in.a][in.b]=1;
            }
        }//s->b
    }
    return -1;
}


int main(){
    while(scanf("%d%d%d",&s,&a,&b)!=EOF&&(s+a+b)!=0){
        if(a<b) swap(a,b);
        if(s&1){
            printf("NO\n");
            continue;
        }
        int ttt=bfs();
        if(ttt==-1)
            printf("NO\n");
        else
            printf("%d\n",ttt);
    }
    return 0;
}