uva 102 - Ecological Bin Packing(暴力)
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题目链接:uva 102 - Ecological Bin Packing
题目大意:有三种灯B,C,G,然后有三个箱子,每个箱子中各装有三种若干个灯,给出每个箱子中各种灯的数量。每次可以从一个箱子中移动一盏灯到另外一个箱子中,问说最少移动几次可以将灯分类,即每一个箱子中只有一种灯。注意每个箱子给出灯的顺序是B,G,C,然而输出是字典序要最小。
解题思路:暴力枚举,维护最小值以及答案即可。
#include <stdio.h>#include <string.h>const int N = 5;const char sign[5] = "BGC";int ans, sum, v[N], t[N], c[N][N], p[N];bool judge() {char s1[N], s2[N];for (int i = 0; i < 3; i++) {s1[i] = sign[t[i]];s2[i] = sign[p[i]];}s1[3] = s2[3] = '\0';return strcmp(s1, s2) < 0;}void dfs(int d, int s) {if (d == 3) {if (s > ans || (s == ans && judge())) {ans = s;memcpy(p, t, sizeof(t));}return ;}for (t[d] = 0; t[d] < 3; t[d]++) {int& u = t[d];if (v[u]) continue;v[u] = 1;dfs(d+1, s+c[d][u]);v[u] = 0;}}bool init() {sum = ans = 0;memset(v, 0, sizeof(v));for (int i = 0; i < 3; i++) {for (int j = 0; j < 3; j++) {if (scanf("%d", &c[i][j]) != 1) return false;sum += c[i][j];}}return true;}int main () {while (init() ) {dfs(0, 0);for (int i = 0; i < 3; i++) printf("%c", sign[p[i]]);printf(" %d\n", sum - ans);}return 0;}
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