1045 Favorite Color Stripe (30)（30 分）

Eva is trying to make her own color stripe out of a given one. She would like to keep only her favorite colors in her favorite order by cutting off those unwanted pieces and sewing the remaining parts together to form her favorite color stripe.

It is said that a normal human eye can distinguish about less than 200 different colors, so Eva's favorite colors are limited. However the original stripe could be very long, and Eva would like to have the remaining favorite stripe with the maximum length. So she needs your help to find her the best result.

Note that the solution might not be unique, but you only have to tell her the maximum length. For example, given a stripe of colors {2 2 4 1 5 5 6 3 1 1 5 6}. If Eva's favorite colors are given in her favorite order as {2 3 1 5 6}, then she has 4 possible best solutions {2 2 1 1 1 5 6}, {2 2 1 5 5 5 6}, {2 2 1 5 5 6 6}, and {2 2 3 1 1 5 6}.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=200) which is the total number of colors involved (and hence the colors are numbered from 1 to N). Then the next line starts with a positive integer M (<=200) followed by M Eva's favorite color numbers given in her favorite order. Finally the third line starts with a positive integer L (<=10000) which is the length of the given stripe, followed by L colors on the stripe. All the numbers in a line are separated by a space.

Output Specification:

For each test case, simply print in a line the maximum length of Eva's favorite stripe.

Sample Input:

6
5 2 3 1 5 6
12 2 2 4 1 5 5 6 3 1 1 5 6

Sample Output:

7


把喜欢的都标记为1，然后裁剪时，只把喜欢的加入序列，对于第i个，找到前面最近的且不是排在他后边的来更新当前的最大长度。所以一开始要标记一下check[a][b]如果是1，表示a在b前，如果是-1,表示a在b后，如果是0就是相等的。
代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
int n,m,l,like_order[201],test[10001];
int check[201][201],iflike[201],dp[10001],ma;
int main() {
    scanf("%d",&n);
    scanf("%d",&m);
    for(int i = 0;i < m;i ++) {
        scanf("%d",&like_order[i]);
        iflike[like_order[i]] = 1;
        for(int j = 0;j < i;j ++) {///标记前后
            check[like_order[j]][like_order[i]] = 1;
            check[like_order[i]][like_order[j]] = -1;
        }
    }
    scanf("%d",&l);
    for(int i = 0;i < l;i ++) {
        scanf("%d",&test[i]);
        if(!iflike[test[i]]) {///如果不是喜欢的直接过了，并且不保存，如果用vector，就直接不加入序列
            i --;
            l --;
            continue;
        }
        int k = i - 1;
        while(k >= 0 && check[test[k]][test[i]] == -1) {
            k --;
        }///只要不是排在后边的  都可以放在前边
        dp[i] ++;///自己本身长度为1
        if(k >= 0)dp[i] += dp[k];///如果 存在就加上他的长度
        ma = max(ma,dp[i]);///更新最大
    }
    printf("%d",ma);
}

这道题如果标记喜欢的颜色的位置，就是求最长非递减子序列。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
int n,m,l,pos[201];
int dp[10001],d,c;
int main() {
    scanf("%d",&n);
    scanf("%d",&m);
    for(int i = 1;i <= m;i ++) {
        scanf("%d",&d);
        pos[d] = i;
    }
    scanf("%d",&l);
    for(int i = 0;i < l;i ++) {
        scanf("%d",&d);
        if(!pos[d])continue;
        if(!c || pos[dp[c - 1]] <= pos[d])dp[c ++] = d;
        else {
            int l = 0,r = c - 1,mid;
            while(l < r) {
                mid = (l + r) / 2;
                if(pos[dp[mid]] <= pos[d])l = mid + 1;
                else r = mid;
            }
            dp[l] = d;
        }
    }
    printf("%d",c);
}