UVa 317 - Hexagon

題目：如圖的19個小六邊形（對邊數字相同）拼成的大六邊形，在3行數字鐘分別選擇三條直線的值，

            使得所有19個六邊形不同（一共有27種選擇），求出直線上數字加和的最大值（相同才有值）。

分析：搜索，枚舉。這裡我是打表計算的。

            首先，使得直線上的數字相同時才能取得最大值否則損失了這條直線的值；

            然後，利用一個程序打表計算所有情況，去掉有重複的小六邊形情況；

                        發現，只有三種選擇情況（三個數字的取值次數：5, 6, 8|5, 7, 7|5, 7, 7）；

            最後，編程枚舉三種排列方式，找到最大值即可。

說明：還有三道題╮(╯▽╰)╭。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;int value[3][3];int list[3][9] = {5, 6, 8, 5, 7, 7, 5, 7, 7,5, 7, 7, 5, 6, 8, 5, 7, 7,5, 7, 7, 5, 7, 7, 5, 6, 8,};int main(){int n;while (~scanf("%d",&n))for (int t = 1; t <= n; ++ t) {for (int i = 0; i < 3; ++ i) {for (int j = 0; j < 3; ++ j) {scanf("%d",&value[i][j]);}}for (int i = 0; i < 3; ++ i) {sort(value[i], value[i]+3);}int max = 0;for (int i = 0; i < 3; ++ i) {int sum = 0;for (int j = 0; j < 3; ++ j) {for (int k = 0; k < 3; ++ k) {sum += list[i][j*3+k] * value[j][k];}}if (max < sum) {max = sum;}}printf("Test #%d\n%d\n\n",t,max);}return 0;}

打表程序：

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <string>#include <set>using namespace std;int map[3][3][3];int size[5] = {3, 4, 5, 4, 3};int main(){set <string>ans;int a[5], b[5], c[5], total = 0;for ( a[0] = 0; a[0] < 3; ++ a[0]) for ( a[1] = 0; a[1] < 3; ++ a[1]) for ( a[2] = 0; a[2] < 3; ++ a[2]) for ( a[3] = 0; a[3] < 3; ++ a[3]) for ( a[4] = 0; a[4] < 3; ++ a[4]) for ( b[0] = 0; b[0] < 3; ++ b[0]) for ( b[1] = 0; b[1] < 3; ++ b[1]) for ( b[2] = 0; b[2] < 3; ++ b[2]) for ( b[3] = 0; b[3] < 3; ++ b[3]) for ( b[4] = 0; b[4] < 3; ++ b[4]) for ( c[0] = 0; c[0] < 3; ++ c[0]) for ( c[1] = 0; c[1] < 3; ++ c[1]) for ( c[2] = 0; c[2] < 3; ++ c[2]) for ( c[3] = 0; c[3] < 3; ++ c[3]) for ( c[4] = 0; c[4] < 3; ++ c[4]) {for (int i = 0; i < 3; ++ i)for (int j = 0; j < 3; ++ j)for (int k = 0; k < 3; ++ k)map[i][j][k] = 0;//第一列 if (map[a[0]][b[0]][c[2]]) continue;map[a[0]][b[0]][c[2]] = 1;if (map[a[0]][b[1]][c[3]]) continue;map[a[0]][b[1]][c[3]] = 1;if (map[a[0]][b[2]][c[4]]) continue;map[a[0]][b[2]][c[4]] = 1;//第二列 if (map[a[1]][b[0]][c[1]]) continue;map[a[1]][b[0]][c[1]] = 1;if (map[a[1]][b[1]][c[2]]) continue;map[a[1]][b[1]][c[2]] = 1;if (map[a[1]][b[2]][c[3]]) continue;map[a[1]][b[2]][c[3]] = 1;if (map[a[1]][b[3]][c[4]]) continue;map[a[1]][b[3]][c[4]] = 1;//第三列 if (map[a[2]][b[0]][c[0]]) continue;map[a[2]][b[0]][c[0]] = 1;if (map[a[2]][b[1]][c[1]]) continue;map[a[2]][b[1]][c[1]] = 1;if (map[a[2]][b[2]][c[2]]) continue;map[a[2]][b[2]][c[2]] = 1;if (map[a[2]][b[3]][c[3]]) continue;map[a[2]][b[3]][c[3]] = 1;if (map[a[2]][b[4]][c[4]]) continue;map[a[2]][b[4]][c[4]] = 1;//第四列 if (map[a[3]][b[1]][c[0]]) continue;map[a[3]][b[1]][c[0]] = 1;if (map[a[3]][b[2]][c[1]]) continue;map[a[3]][b[2]][c[1]] = 1;if (map[a[3]][b[3]][c[2]]) continue;map[a[3]][b[3]][c[2]] = 1;if (map[a[3]][b[4]][c[3]]) continue;map[a[3]][b[4]][c[3]] = 1;//第五列 if (map[a[4]][b[2]][c[0]]) continue;map[a[4]][b[2]][c[0]] = 1;if (map[a[4]][b[3]][c[1]]) continue;map[a[4]][b[3]][c[1]] = 1;if (map[a[4]][b[4]][c[2]]) continue;map[a[4]][b[4]][c[2]] = 1;int count[3][3];for (int i = 0; i < 3; ++ i)for (int j = 0; j < 3; ++ j)count[i][j] = 0;//printf("  a:");for (int i = 0; i < 5; ++ i) {printf("%d, ",a[i]);count[0][a[i]] += size[i];}sort(count[0], count[0]+3);//printf("  b:");for (int j = 0; j < 5; ++ j) {printf("%d, ",b[j]);count[1][b[j]] += size[j];}sort(count[1], count[1]+3);//printf("  c:");for (int k = 0; k < 5; ++ k) {printf("%d, ",c[k]);count[2][c[k]] += size[k];}sort(count[2], count[2]+3);printf(" : ");string str = "000000000";for (int i = 0; i < 3; ++ i)for (int j = 0; j < 3; ++ j) {printf("%d ",count[i][j]);str[i*3+j] += count[i][j];}puts("");total ++;if (ans.count(str) > 0) {;}else {ans.insert(str);}}printf("total = %d\n",total);// 输出取值的组合 set<string>::reverse_iterator rit;for (rit = ans.rbegin(); rit != ans.rend(); ++ rit) {cout << *rit << endl;}getchar();return 0;}