快速切题 poj3414 Pots

Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10042		Accepted: 4221		Special Judge

Description

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

1. FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
2. DROP(i)      empty the pot i to the drain;
3. POUR(i,j)    pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).

Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

Input

On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).

Output

The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

Sample Input

3 5 4

Sample Output

6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)

应用时:10min
实际用时:57min
原因:六种操作都是不可逆转的.
思路:时间非常充裕到不需要建反边,数据太小

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=101;
int n,m;
typedef unsigned long long ull;
int A,B,C;
int vis[maxn][maxn];
int ans[maxn][maxn][maxn*maxn];
struct node{
    int a,b;
    node (int  ta,int tb):a(ta),b(tb){}
};
void printop(int op){
    switch(op){
    case 0:
        puts("FILL(1)");
        break;
    case 1:
        puts("FILL(2)");
        break;
    case 2:
        puts("POUR(1,2)");
        break;
    case 3:
        puts("POUR(2,1)");
        break;
    case 4:
        puts("DROP(1)");
        break;
    case 5:
        puts("DROP(2)");
        break;
    }
}
void op(int &a,int &b,int op){
    switch(op){
    case 0:
        a=A;
        break;
    case 1:
        b=B;
        break;
    case 2:
        if(b+a<=B){
            b+=a;
            a=0;
        }
        else {
            a-=B-b;
            b=B;
        }
        break;
    case 3:
        if(b+a<=A){
            a+=b;
            b=0;
        }
        else {
            b-=A-a;
            a=A;
        }
        break;
    case 4:
        a=0;
        break;
    case 5:
        b=0;
        break;
    }
}
void bfs(){
    queue <node> que;
    que.push(node(0,0));
    vis[0][0]=0;
    while(!que.empty()){
        node tp=que.front();que.pop();
        int ta=tp.a;
        int tb=tp.b;
        if(tp.a==C||tp.b==C){
            printf("%d\n",vis[tp.a][tp.b]);
            for(int i=0;i<vis[ta][tb];i++){
                int op=ans[ta][tb][i];
                printop(op);
            }
            return ;
        }
        for(int i=0;i<6;i++){
            int ta=tp.a;
            int tb=tp.b;
            op(ta,tb,i);
            if(vis[ta][tb]==-1){
                vis[ta][tb]=vis[tp.a][tp.b]+1;
                for(int j=0;j<vis[tp.a][tp.b];j++){
                    ans[ta][tb][j]=ans[tp.a][tp.b][j];
                }
                ans[ta][tb][vis[tp.a][tp.b]]=i;
                que.push(node(ta,tb));
            }
        }
    }
    puts("impossible");
}
int main(){
    scanf("%d%d%d",&A,&B,&C);
        memset(vis,-1,sizeof(vis));
        bfs();
    return 0;
}