三個水杯

描述

給出三個水杯，大小不一，並且只有最大的水杯的水是裝滿的，其余兩個為空杯子。三個水杯之間相互倒水，並且水杯沒有標識，只能根據給出的水杯體積來計算。現在要求你寫出一個程序，使其輸出使初始狀態到達目標狀態的最少次數。

輸入

第一行一個整數N(0<N<50)表示N組測試數據
接下來每組測試數據有兩行，第一行給出三個整數V1 V2 V3 (V1>V2>V3 V1<100 V3>0)表示三個水杯的體積。
第二行給出三個整數E1 E2 E3 （體積小於等於相應水杯體積）表示我們需要的最終狀態

輸出

每行輸出相應測試數據最少的倒水次數。如果達不到目標狀態輸出-1

樣例輸入

2
6 3 1
4 1 1
9 3 2
7 1 1

樣例輸出

3
-1

來源

經典題目

廣度優先搜索，貌似可以用A*，可惜不怎麽會，以後試試。

# include <stdio.h>  
# include <stdlib.h>  
# include <string.h>  
# include <string>  
  
void check (int queue[], int x, int &tail) {  
    for (int i = 0; i <= tail; ++ i) {  
        
 
if (queue[i] == x) {  
            return;  
        }  
    }  
    ++ tail;  
    //printf ("%d -> ", x);  
    queue[tail] = x;  
    return;  
}  
  
int getans(int to[], int i) {  
    if (i == to[1] * 10000 + to[2] * 100 + to[3]) return 1;  
    return 0;  
}  
  
int daoshui(int queue[], int &a, int
 
 &b, int to, int from[]) {  
    if (a) {  
        int t = from[to] - b;  
        if (t > a) {  
            b += a;  
            a = 0;  
            return 1;  
        }  
        else if (t) {  
            a -= t;  
            b += t;  
            return 1;  
        }  
    }  
    return 0;  
}  
  
int main () {  
    int n;  
    scanf ("%d", &n);  
    while (n --) {  
        int from[4];  
        int to[4];  
        scanf ("%d %d %d %d %d %d", &from[1], &from[2], &from[3], &to[1], &to[2], &to[3]);  
        int head = 0;  
        int tail = 0;  
        int step = 0;   
        int queue[100000];  
        queue[0] = from[1] * 10000;  
        int flag = 0;  
        if (queue[0] == to[1] * 10000 + to[2] * 100 + to[3]) {printf("0\n");continue;}  
        while (head <= tail) {  
            ++ step;  
            //printf("******************************%d************************************\n", step);  
            int size = tail - head;  
            for (int i = head; i <= head + size; ++ i) {  
                int a, b, c;  
                //printf ("|%d|", queue[i]);  
                a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;  
                if (daoshui(queue, a, b, 2, from)) {  
                    //printf ("(a -> b)");  
                    check(queue, a * 10000 + b * 100 + c, tail);  
                }  
                a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;  
                if (daoshui(queue, a, c, 3, from)) {  
                    //printf ("(a -> c)");  
                    check(queue, a * 10000 + b * 100 + c, tail);  
                }  
                a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;  
                if (daoshui(queue, b, a, 1, from)) {  
                    //printf ("(b -> a)");  
                    check(queue, a * 10000 + b * 100 + c, tail);  
                }  
                a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;  
                if (daoshui(queue, b, c, 3, from)) {  
                    //printf ("(b -> c)");  
                    check(queue, a * 10000 + b * 100 + c, tail);  
                }  
                a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;  
                if (daoshui(queue, c, a, 1, from)) {  
                    //printf ("(c -> a)");  
                    check(queue, a * 10000 + b * 100 + c, tail);  
                }  
                a = queue[i] / 10000; b = queue[i] % 10000 / 100; c = queue[i] % 100;  
                if (daoshui(queue, c, b, 2, from)) {  
                    //printf ("(c -> b)");  
                    check(queue, a * 10000 + b * 100 + c, tail);  
                }  
            }  
            //printf ("\n****************************************************************");  
            head += size + 1;  
            for (int i = head; i <= tail; ++ i) {  
                if (getans(to, queue[i])) {  
                    printf ("%d\n", step);  
                    flag = 1;  
                    break;  
                }  
            }  
            if (flag) break;  
        }  
        //printf("\n");  
        if (!flag) printf ("-1\n");  
    }  
    return 0;  
}  

三個水杯