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1401: Sherlock and Moriarty II

时间限制: 1 Sec  内存限制: 128 MB
提交: 89  解决: 31
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题目描述

Sherlock Holmes is an excellent detective and Moriarty is the big villain. One day Moriarty kidnaps Sherlock's
boyfriend ---- Jhon Watson. Of course,Sherlock goes to save him. As sherlock has found the password in the previous
problem.When he finds Jhon, there is a time bomb. Sherlock has to defuse the bomb.
The bomb has a special coded lock,the lock has two sequence A and B, Sherlock has to find a new Sequence C , C is
both the subsequence of A and B, and for each i<j C[i]<C[j]. The length of C should be as long as possible.The length
of sequence C is the password.

输入

There are several cases.
For each case,the first line is N and M (1<=n,m<=100)
And the is sequence A and B , length(A)=N length(B)=M （0<= A[i],B[i]<=50）

输出

The length of sequence C.

样例输入

5 5
1 2 0 4 5
1 0 4 5 2

样例输出

3

这是最复杂

#include<stdio.h>
#include<string.h>
struct node{ int max, len; };
int a[101];
int b[101];
node c[101][101];
int asize, bsize;
int main(){
	//freopen("in.txt", "r", stdin);
	while (~scanf("%d%d", &asize, &bsize)){
		int i, j,m,n;
		for (i = 0; i < asize; i++)
			scanf("%d", &a[i]);
		for (i = 0; i < bsize; i++)
			scanf("%d", &b[i]);
		memset(c, 0, sizeof(c));
		for (i = 0; i < asize;i++)
		for (j = 0; j < bsize; j++)
		{
			if (a[i] != b[j])continue;
			c[i][j].len = 1;
			c[i][j].max = a[i];
			for (m = 0; m < i;m++)
			for (n = 0; n < j; n++)
			if (c[m][n].max<a[i] && 1 + c[m][n].len>c[i][j].len)
				c[i][j].len = 1 + c[m][n].len;
		}
		m = 0;
		for (i = 0; i < asize;i++)
		for (j = 0; j < bsize; j++){
			if (c[i][j].len>m)m = c[i][j].len;
		}
		printf("%d
", m);
	}
	return 0;
}