题目十五:分配问题,一共有21个筐,7个筐有满筐货物,7个筐有半筐货物,7个空筐,将筐和货物平均分给三个人,求所有的分配方案
/*
题目十五: 分配问题,一共有21个筐,7个筐有满筐货物,7个筐有半筐货物,7个空筐,将筐和货物平均分给三个人,求所有的分配方案
分析:平均每人需要分到7个筐和3.5筐货物,也就是说每人至少得有奇数个半筐货物,且最多有五筐半筐,根据之前的
题目可以采用穷举法尝试
穷举法过于麻烦,观察到本题有两个限制条件
分配方式可以为:
x y 7-x-y
a b 7-a-b
7-a-x 7-y-b x+y+a+b-7
可得改进后的穷举法
利用数组做,可得一个3*3方阵,数组的每行,每列的和为7;
行数中满筐+半筐*0.5=3.5;
每人的满筐数不得超过3
每人的半筐数至少为一,且为奇数
注意:二维数组传参数时,在函数中第二个参数必须给出
例如:int test_all(int all[][3]){}为正确写法
*/
#include<iostream>
using namespace std;
int test_true(int all,int bin){
float total = 3.5;
if(all+0.5*bin == total){
return 1;
}else{
return 0;
}
}
int test_all(int a[3][3]){
int i,j,total_row = 0,total_len = 0;
float total_shop = 0.0;
for(i = 0;i<3;i++){
total_shop += (a[i][0]+0.5*a[i][1]);
for(j = 0;j<3;j++){
total_row += a[i][j];
total_len += a[j][i];
if (total_row != 7 || total_len !=7 || total_shop!=3.5){
return 0;
}else{
total_row = 0;
total_len = 0;
total_shop = 0.0;
}
}
}
return 1;
}
int main(){
//题目十五
//直接穷举麻烦且循环嵌套太多,不建议使用
/*
int basket_bin_1,basket_bin_2,basket_bin_3;
float shop_1 = 0.0,shop_2 = 0.0,shop_3 = 0.0;
int basket_1 = 0,basket_2 = 0,basket_3 = 0;
int i,j,k;
for(basket_bin_1 = 1;basket_bin_1 <= 5;basket_bin_1+=2){
basket_1 = 0;
shop_1 = 0.5*basket_bin_1;
if(basket_2 != 0){
for(i=0;i<=7-basket_bin_1;i++){
if((7-(i+basket_bin_3+basket_bin_2+basket_bin_1))+shop_1==3.5){
cout<<"A: "<<i<<" "<<basket_bin_1<<" "<<7-i-basket_bin_1
<<"B: "<<j<<" "<<basket_bin_2<<" "<<7-j-basket_bin_2
<<"C: "<<k<<" "<<basket_bin_3<<" "<<7-k-basket_bin_3<<endl;
break;
}
}
}
for(basket_bin_2 = 1;basket_bin_2 <= 7-basket_bin_1;basket_bin_2+=2){
basket_bin_3 = 7-basket_bin_1-basket_bin_2;
basket_2 = 0;
shop_2 = 0.5*basket_bin_2;
if(basket_3 != 0){
for(j=0;j<=7-basket_bin_2;j++){
if((7-(j+basket_bin_3+basket_bin_2))+shop_2==3.5){
basket_2 = basket_bin_2;
break;
}
}
}
basket_3 = 0;
shop_3 = 0.5*basket_bin_3;
for(k = 0;k<=7-basket_bin_3;k++){
if(((7-(k+basket_bin_3))+shop_3)==3.5){
basket_3 = basket_bin_3;
break;
}
}
}
}
*/
//题目十五,改进后的穷举法
int i,j,all_i,all_j;
for(i = 1;i<=5;i+=2){
for(j = 1;j<=5;j++){
if(i+j > 7)break;
for(all_i = 0;all_i<=3;all_i++){
for(all_j = 0;all_j<=3;all_j++){
if(all_i+all_j > 7)break;
if(test_true(all_i,i)&&test_true(all_j,j)&&test_true(7-all_i-all_j,7-i-j)){
cout<<"A: 满:"<<all_i<<" 半:"<<i<<" 空:"<<7-all_i-i<<endl
<<"B: 满:"<<all_j<<" 半:"<<j<<" 空:"<<7-all_j-j<<endl
<<"C: 满:"<<7-all_i-all_j<<" 半:"<<7-i-j<<" 空:"<<all_i+i+all_j+j-7<<endl<<endl;
}
}
}
}
}
//题目十五,数组方式
/*
int all[3][3];
int i,j;
int a,b;
int x,y;
for(i = 0;i<=3;i++){
for(j = 0;j<=3;j++){
for(a = 1;a<=5;a+=2){
for(b = 1;b<=5;b+=2){
if(a+b>7)break;
if(i+j+b+a-7 < 0)break;
all[0][0] = i;
all[0][1] = a;
all[0][2] = 7-i-a;
all[1][0] = j;
all[1][1] = b;
all[1][2] = 7-j-b;
all[2][0] = 7-i-j;
all[2][1] = 7-a-b;
all[2][3] = i+j+b+a-7;
if(test_all(all)){
for(x = 0;x<3;x++){
for(y = 0;y<3;y++){
cout<<" "<<all[x][y]<<" ";
}
cout<<endl;
}
cout<<endl;
}
}
}
}
}
*/
return 0;
}
这其中有不合适或者不正确的地方欢迎指正,我的QQ号码:2867221444(乔金明),谢谢,也可以相互交流下,备注信息随意,只要能看得出是开发者或者学习者即可。