Codeforces Beta Round #10 D. LCIS

题目链接：

http://www.codeforces.com/contest/10/problem/D

D. LCIS

time limit per test:1 second
memory limit per test:256 megabytes

问题描述

This problem differs from one which was on the online contest.

The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.

The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 ≤ i1 < i2 < ... < ik ≤ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.

You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.

输入

The first line contains an integer n (1 ≤ n ≤ 500) — the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] — elements of the first sequence. The third line contains an integer m (1 ≤ m ≤ 500) — the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] — elements of the second sequence.

输出

In the first line output k — the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.

样例

input

7

2 3 1 6 5 4 6

4

1 3 5 6

output

3

3 5 6

题意

求最长公共递增子序列，并输出一个最优解。

题解

dp[j]表示第一个串的前i个和第二个串的前j个的以b[j]结尾的公共最长上升子序列的长度。

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 555;

int dp[maxn];
int a[maxn], b[maxn];
int pre[maxn];
int n, m;

void print(int i) {
	if (!i) return;
	print(pre[i]);
	printf("%d ", b[i]);
}

void init() {
	memset(dp, 0, sizeof(dp));
	memset(pre, 0, sizeof(pre));
}

int main() {
	init();
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	scanf("%d", &m);
	for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
	int ans = 0, st = 0;
	for (int i = 1; i <= n; i++) {
		int pos = 0;
		for (int j = 1; j <= m; j++) {
			if (b[j]<a[i] && dp[pos]<dp[j]) {
				pos = j;
			}
			else if (a[i] == b[j]) {
				dp[j] = dp[pos] + 1;
				pre[j] = pos;
			}
			//这样边扫边记会wa，还没找到原因。。
			//if (ans<dp[j]) {
			//	ans = dp[j], st = j;
			//}
		}
	}
	for (int i = 1; i <= m; i++) {
		if (dp[i] > ans) {
			ans = dp[i], st = i;
		}
	}
	printf("%d\n", ans);
	print(st);
	return 0;
}

再一发：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=555;
const int maxm=555;

int a[maxn],b[maxn];
int dp[maxn][maxn],pre[maxn][maxn];

int main() {
    int n;
    scf("%d",&n);
    for(int i=1;i<=n;i++) scf("%d",&a[i]);
    int m;
    scf("%d",&m);
    for(int i=1;i<=m;i++) scf("%d",&b[i]);

    clr(dp,0);
    clr(pre,0);

    for(int i=1;i<=n;i++){
        int ma=0,p=0;
        for(int j=1;j<=m;j++){
            dp[i][j]=dp[i-1][j];
            pre[i][j]=pre[i-1][j];
            if(a[i]==b[j]){
                if(dp[i][j]<ma+1){
                    dp[i][j]=ma+1;
                    pre[i][j]=p;
                }
            }else if(b[j]<a[i]){
                if(ma<dp[i-1][j]){
                    ma=dp[i-1][j];
                    p=j;
                }
            }
        }
    }

    int ans=0,pos=0;
    for(int i=1;i<=m;i++){
        if(ans<dp[n][i]){
            ans=dp[n][i];
            pos=i;
        }
    }
    VI lis;
    lis.pb(pos);
    int p=pre[n][pos];
    while(p){
        lis.pb(p);
        p=pre[n][p];
    }
    reverse(lis.begin(),lis.end());
    printf("%d\n",ans);
    if(ans==0) return 0;
    rep(i,0,lis.sz()-1) prf("%d ",b[lis[i]]);
    prf("%d\n",b[lis[lis.sz()-1]]);
    return 0;
}

//end-----------------------------------------------------------------------

/*
3 1 2 3
3 3 4 5
*/