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1. Title
You are installing a billboard and want it to have the highest height.
This billboard will have two steel brackets, one on each side. The height of each steel support must be equal .
You have a bunch of steel rods that can be welded together.
For example, if the lengths of the steel bars are 1, 2 and 3, they can be welded together to form a 6-length bracket.
Returns the maximum possible installation height of the billboard . If the billboard cannot be installed, please return 0.
示例 1：
输入：[1,2,3,6]
输出：6
解释：我们有两个不相交的子集 {
1,2,3} 和 {
6}，它们具有相同的和 sum = 6。
示例 2：
输入：[1,2,3,4,5,6]
输出：10
解释：我们有两个不相交的子集 {
2,3,5} 和 {
4,6}，它们具有相同的和 sum = 10。
示例 3：
输入：[1,2]
输出：0
解释：没法安装广告牌，所以返回 0。
提示：
0 <= rods.length <= 20
1 <= rods[i] <= 1000
钢筋的长度总和最多为 5000
Source: LeetCode (LeetCode) Link: https://leetcode-cn.com/problems/tallest-billboard
Copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.
2. Problem solving
dp[i][j]Indicates the maximum height that can be formed when the i steel bar is processed and the gap between the two sides is j
class Solution {
public:
int tallestBillboard(vector<int>& rods) {
if(rods.size() <= 1)
return 0;
int dp[20][5001], n = rods.size();
int total = accumulate(rods.begin(), rods.end(), 0);
// dp[i][j] 表示处理完 i 支架，两边差距为 j 时，可以组成的最大高度
memset(dp, -1, sizeof(dp));
dp[0][rods[0]] = dp[0][0] = 0;
for(int i = 1; i < rods.size(); i++)
{
//样本维度
for(int j = 0; j <= total; ++j)
{
if(dp[i-1][j] == -1)//上一行状态不存在
continue;
// 当前钢筋不要，丢弃
dp[i][j] = max(dp[i][j], dp[i-1][j]);
// 当前钢筋，加在长的一边
dp[i][j+rods[i]] = max(dp[i][j+rods[i]], dp[i-1][j]);
// 当前钢筋，加在短的一边，现在高度差为 abs(j-rods[i])
dp[i][abs(j-rods[i])] = max(dp[i][abs(j-rods[i])], dp[i-1][j]+min(j, rods[i]));
}
}
// 返回高度差为 0 的情况
return dp[n-1][0]==-1 ? 0 : dp[n-1][0];
}
};
40 ms 8.1 MB
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