Hdu1423 Greatest Common Increasing Subsequence

Description

Input

Output

Sample Input

Sample Output

做法一：复杂度\(O(N^3)\)。

我们用\(f_{i,j}\)表示\(A\)数组前\(i\)位，B数组前\(j\)位的最长公共上升子串有多长，但是只有\(A_i = B_j\)的时候有意义。dp方程不难想，但是仔细一算复杂度这是\(O(n^4)\)的，为此我们需要加上一些优化。

我们先枚举\(i\)，我们可以维护一个数组\(P\)。其中\[P_j = \max \{ f_{k,j} \},k < i\]

然后dp方程就可以变成\[f_{i,j} = \max\{ P_k+1 \},k < j \; and \; B_k < B_j\]

\(P\)每次dp一个\(i\)都可以\(O(N)\)维护出来。

做法二：复杂度\(O(N^2)\)

令\(g_{i,j} = \max \{ f_{k,j},k \le i \}\)，然后我们想想怎么用\(g\)直接来dp。

令\(ma = \max \{ f_{i-1,k},k < j \; and \; A_i > B_k \}\)

首先\[g_{i,j} = g_{i-1,j}\]

然后若\(A_i = B_j\)，则\[g_{i,j} = \max\{ g_{i,j},ma+1 \}\]

最后\[ans = \max \{ g_{N,i} \},1 \le i \le M\]

代码1：

#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;

const int maxn = 510,inf = 1<<30;
int f[maxn][maxn],A[maxn],B[maxn],P[maxn],T,N,M,ans;

int main()
{
freopen("1423.in","r",stdin);
freopen("1423.out","w",stdout);
scanf("%d",&T);
while (T--)
{
scanf("%d",&N); ++N; A[1] = 0;
for (int i = 2;i <= N;++i) scanf("%d",A+i);
scanf("%d",&M); ++M; B[1] = 0;
for (int i = 2;i <= M;++i) scanf("%d",B+i);
for (int i = 1;i <= N;++i) for (int j = 1;j <= M;++j) f[i][j] = -inf;
for (int i = 1;i <= M;++i) P[i] = -inf;
ans = f[1][1] = P[1] =0;
for (int i = 2;i <= N;++i)
{
for (int j = 2;j <= M;++j)
{
if (A[i] != B[j]) continue;
for (int k = 1;k < j;++k) if (B[k] < A[i]) f[i][j] = max(f[i][j],P[k]+1);
ans = max(ans,f[i][j]);
}
for (int j = 2;j <= M;++j) P[j] = max(P[j],f[i][j]);
}
printf("%d\n",ans); if (T) puts("");
}
fclose(stdin); fclose(stdout);
return 0;
}

代码2：

#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;

const int maxn = 510,inf = 1<<30;
int f[maxn][maxn],A[maxn],B[maxn],P[maxn],T,N,M,ans;

int main()
{
freopen("1423.in","r",stdin);
freopen("1423.out","w",stdout);
scanf("%d",&T);
while (T--)
{
scanf("%d",&N); ++N; A[1] = 0;
for (int i = 2;i <= N;++i) scanf("%d",A+i);
scanf("%d",&M); ++M; B[1] = 0;
for (int i = 2;i <= M;++i) scanf("%d",B+i);
for (int i = 1;i <= N;++i) for (int j = 1;j <= M;++j) f[i][j] = -inf;
for (int i = 1;i <= M;++i) P[i] = -inf;
ans = f[1][1] = P[1] =0;
for (int i = 2;i <= N;++i)
for (int j = 2,ma = 0;j <= M;++j)
{
f[i][j] = f[i-1][j];
if (A[i] == B[j]) f[i][j] = ma+1;
if (A[i] > B[j]&&ma < f[i-1][j]) ma = f[i-1][j];
}
for (int i = 1;i <= M;++i) ans = max(ans,f
[i]);
printf("%d\n",ans); if (T) puts("");
}
fclose(stdin); fclose(stdout);
return 0;
}