poj 3414( 搜索 )

You are given two pots,having the volume ofAandBliters respectively. The following operations can be performed:

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

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

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

The first line of the output must contain the length of the sequence of operationsK. The followingKlines 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’.

#include <stdio.h>
#include <string.h>
#define N 101

struct Way {
    int prex;
    int prey;
    int kind;
} way[N][N];
struct Point{
    int x;
    int y;
} queue[N*N*2],c,r;

int answer[N*N];
int index = -1;
int visit[N][N];
int va,vb,vc;

void show(){
    int i;
    char temp[6][10] = { "FILL(1)","FILL(2)","DROP(1)","DROP(2)","POUR(1,2)","POUR(2,1)"};
    printf("%dn",index+1);
    for( i = index; i >= 0; i-- ){
        printf("%sn",temp[answer[i]]);
    }
}

int bfs(){
    int top = 0,end = 0,i;
    index = -1;
    memset(visit,sizeof(visit));
    visit[0][0] = 1;
    queue[0].x = 0;
    queue[0].y = 0;

    while(top <= end){
        c = queue[top];
        for( i = 0; i < 6; i++){
            switch( i ){
                case 0: r.x = va; r.y = c.y;      break; //fill 1
                case 1: r.x = c.x; r.y = vb;      break; //fill 2
                case 2: r.x = 0; r.y = c.y;       break; //drop 1
                case 3: r.x = c.x; r.y = 0;       break; //drop 2
                case 4:                                  //pour 1 2
                        if( c.x + c.y <= vb) {
                            r.x = 0;
                            r.y = c.x + c.y;
                        } else {
                            r.x = c.x - (vb - c.y);
                            r.y = vb;
                        }
                        break;
                case 5:                                  //pour 2 1
                        if( c.x + c.y <= va){
                            r.x = c.x + c.y;
                            r.y = 0;
                        } else {
                            r.x = va;
                            r.y = c.y - (va - c.x);
                        }

                        break;
            }
            if( visit[r.x][r.y] == 0){
                if( r.x == vc || r.y == vc){
                    answer[++index] = i;
                    int nx = c.x;
                    int ny = c.y;
                    int prex,prey;
                    while( nx != 0 || ny != 0){
                        answer[++index] = way[nx][ny].kind;
                        prex = way[nx][ny].prex;
                        prey = way[nx][ny].prey;
                        nx = prex;
                        ny = prey;
                    }
                    return 1;
                } else {
                    visit[r.x][r.y] = 1;
                    queue[++end] = r;
                    way[r.x][r.y].prex = c.x;
                    way[r.x][r.y].prey = c.y;
                    way[r.x][r.y].kind = i;
                }
            }
        }
        top++;
    }
    return -1;
}


int main()
{
    while(scanf("%d%d%d",&va,&vb,&vc) != EOF){
        if(bfs() == -1){
            printf("impossiblen");
        } else {
            show();
        }
    }

    return 0;
}