Prince and Princess UVA - 10635

最新推荐文章于 2020-05-28 14:18:17 发布

原创最新推荐文章于 2020-05-28 14:18:17 发布 · 270 阅读

0 ·

CC 4.0 BY-SA版权

思维专栏收录该内容

24 篇文章

订阅专栏

本文介绍了一种寻找两个序列的最长公共子序列的有效算法。通过将问题转化为最长上升子序列问题，利用动态规划思想实现。文章包含完整的C++代码实现，并详细解释了算法步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

找出两个序列的最长公共子序列

因为每个序列中每个数只出现一次
所以可以先记录每个数字在第一个序列中出现的位置
再把第二个序列的数字换成这个数字在第一个序列中出现的位置
这样求一个最长上升子序列就是答案

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ansn() printf("%d\n",ans)
#define lansn() printf("%lld\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pll pair<long long,long long>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
//const ll mod = 1000000007 ;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
//const ll infl = 100000000000000000;//1e17
const int maxn=  1e5+20;
const int maxm = 1e5+20;
//muv[i]=(p-(p/i))*muv[p%i]%p;
int in(int &ret) {
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-1:1;
    ret = (c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
    ret *=sgn;
    return 1;
}

int a[maxn];
int b[maxn];
int c[maxn];
int main() {
#ifdef LOCAL
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
#endif // LOCAL

    int t;
    sd(t);
    r1(cas,t) {
        printf("Case %d: ",cas);
        int n,p,q;
        sddd(n,p,q);
        int sz = n*n;
        for(int i=1; i<=sz; ++i)a[i] = inf;
        for(int i=1; i<=p+1; ++i) {
            int x;
            sd(x);
            a[x] = i;
        }
        for(int i=1; i<=q+1; ++i) {
            int x;
            sd(x);
            b[i] = a[x];
        }
        int len = 1;
        c[1] = b[1];
//        for(int i=1; i<=q+1; i++)printf("%d%c",b[i]," \n"[i==q+1]);
        for(int i=2; i<=q+1; ++i) {
            if(b[i]==inf)continue;
            if(b[i]>c[len])c[++len] = b[i];
            else {
                int l = 1 , r = len;
                int pos = 0;
                while(l<=r) {
                    int mid = (l+r)>>1;
                    if(b[i]<c[mid])r = mid - 1, pos = mid;
                    else l = mid + 1;
                }
                c[pos] = b[i];
            }
        }
//        for(int i=1; i<=len; i++)printf("%d%c",c[i]," \n"[i==len]);

        int ans = len;
        ansn();
    }
    return 0;
}