UVA 10635 Prince and Princess LCS转LIS

题意：求最长公共子序列

正常的算法O(n^2)不够快，但因为这题数据特殊，数组中的数各不相同，可以转化成求最大上升子序列

比如样例中A{1 7 5 4 8 3 9}， B{1 4 3 5 6 2 8 9}，将数字重新映射编号map[ai]=i;

ai=map[ai];bi=map[bi];

变成A{1,2,3,4,5,6,7}  B{1,4,6,3,0,0,5,7} 问题便变成了求B的LIS可以在0（nlogn）求出

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define pb(a) push_back(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r

void debug()
{
#ifdef ONLINE_JUDGE
#else
    freopen("d:\\in.txt","r",stdin);
    freopen("d:\\out1.txt","w",stdout);
#endif
}
char getch()
{
    char ch;
    while((ch=getchar())!=EOF)
    {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}

struct point
{
    int x,y,t;
    point() {}
    point (int a,int b,int e)
    {
        x=a;
        y=b;
        t=e;
    }
    bool operator < (point a) const
    {
        return t>a.t;
    }
};
int da[70000];
int dp[70000];
int lis(int n)              //LIS的O(nlogn)算法
{
    int maxx=1;
    dp[1]=da[1];
    for(int i=2;i<=n;i++)
    {
        if(da[i]>=dp[maxx])
        {
            dp[++maxx]=da[i];
        }
        else
        {
            int v=upper_bound(dp+1,dp+1+maxx,da[i])-dp;
            dp[v]=da[i];
        }
    }
    return maxx;
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        map<int,int> ma;
        int n,a,b;
        scanf("%d%d%d",&n,&a,&b);
        for(int i=1;i<=a+1;i++)
        {
            int x;
            scanf("%d",&x);
            ma[x]=i;
        }
        int k=0;
        for(int i=1;i<=b+1;i++)
        {
            int x;
            scanf("%d",&x);
            int v=ma[x];
            if(v!=0)//0表示没出现，没必要加进去了
                da[++k]=v;
        }
        int num=lis(k);
        printf("Case %d: %d\n",ca,num);
    }
    return 0;
}