Why Did the Cow Cross the Road II - UPCOJ 3438 - 动态规划DP

题目：

题目描述
Farmer John raises N breeds of cows (1≤N≤1000), conveniently numbered 1…N. Some pairs of breeds are friendlier than others, a property that turns out to be easily characterized in terms of breed ID: breeds a and b are friendly if |a−b|≤4, and unfriendly otherwise.
A long road runs through FJ’s farm. There is a sequence of N fields on one side of the road (one designated for each breed), and a sequence of N fields on the other side of the road (also one for each breed). To help his cows cross the road safely, FJ wants to draw crosswalks over the road. Each crosswalk should connect a field on one side of the road to a field on the other side where the two fields have friendly breed IDs (it is fine for the cows to wander into fields for other breeds, as long as they are friendly). Each field can be accessible via at most one crosswalk (so crosswalks don’t meet at their endpoints).

Given the ordering of N fields on both sides of the road through FJ’s farm, please help FJ determine the maximum number of crosswalks he can draw over his road, such that no two intersect.

输入
The first line of input contains N. The next N lines describe the order, by breed ID, of fields on one side of the road; each breed ID is an integer in the range 1…N. The last N lines describe the order, by breed ID, of the fields on the other side of the road. Each breed ID appears exactly once in each ordering.

输出
Please output the maximum number of disjoint “friendly crosswalks” Farmer John can draw across the road.

样例输入

6
1 2 3 4 5 6
6 5 4 3 2 1

样例输出

5

题意：

  给你一个n，然后是2*n个点，对应a[n]和b[n]两个数组，如果abs(a[i]-b[j])<=4，那么就可以将这两个点连起来。问，在连线不交叉的情况下，最多可以连多少根线。

思路：

  脑补了一下连线不能交叉的样子，像是代码归并对比的那样，于是想到了最长公共子序列。

实现：

#include 
using namespace std;
int n,x[1007],y[1007],dp[1007][1007];
int main() {ios_base::sync_with_stdio(false);cin.tie(nullptr);while(memset(dp,0,sizeof dp), cin >> n) {for(int i=1 ; i<=n ; i++) cin >> x[i];for(int i=1 ; i<=n ; i++) cin >> y[i];for(int i=1 ; i<=n ; i++) {for(int j=1 ; j<=n ; j++) {if(abs(x[i]-y[j]) <= 4) dp[i][j] = dp[i-1][j-1]+1;else dp[i][j] = max(dp[i][j-1],dp[i-1][j]);}}cout << dp[n][n] << '\n';}return 0;
}

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