uva 10635-Prince and Princess 【LIS 求解 LCS】

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题意：给定两个分别具有q+1个和p+1个元素的序列，序列中的元素均不相同且均在1-n*n范围内。让你求解两个序列LCS的长度。

思路：n最大为250，n*n = 250 * 250 = 62500，LCS的O(n^2)算法会TLE。

先将一个序列里的元素建立val-id的映射fp，在另一个序列用fp[val]来替换val，然后求解LIS即可，为了方便求解，可以将fp[val] == 0的元素去掉。这样时间复杂度O(nlogn)。

AC代码:

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (250*250+10)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;int b[MAXN], s[MAXN];map<int, int> fp;int main(){    int t, n, p, q, kcase = 1;    Ri(t);    W(t)    {        Ri(n); Ri(p); Ri(q);        p++; q++;        int id = 0; fp.clear();        int a;        for(int i = 0; i < p; i++)        {            Ri(a);            fp[a] = ++id;        }        int top = 0;        for(int i = 0; i < q; i++)        {            Ri(a);            if(fp[a])                b[top++] = fp[a];        }        int len = 1; s[0] = -1;        for(int i = 0; i < top; i++)        {            s[len] = INF;            int j = lower_bound(s, s+len, b[i]) - s;            if(j == len)                len++;            s[j] = b[i];        }        printf("Case %d: %d\n", kcase++, len-1);    }    return 0;}

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