PAT (Advanced Level) 1045 Favorite Color Stripe (DP)

版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/w419387229/article/details/82143505

 dp[i][j]表示fav中第i个数匹配到stripe第j个数时,所能获得的最大长度。 

由于同样的数字可以重复,因此状态转移方程:dp[i][j] = max(dp[i][j-1], dp[i-1][j]) + k, (k = fav[i]==stripe[j] ? 1 : 0)

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
int n, m, l;
int fav[205];
int stripe[10005];
int dp[205][10005];
 
int main(){
	scanf("%d%d",&n,&m);
	for(int i = 1; i <= m; ++i)
		scanf("%d",&fav[i]);
	scanf("%d",&l);
	for(int i = 1; i <= l; ++i)
		scanf("%d",&stripe[i]);
	memset(dp,0,sizeof(dp));
	for(int i = 1; i <= m; ++i){
		for(int j = 1; j <= l; ++j){
			int maxn = max(dp[i][j-1],dp[i-1][j]);
			if(stripe[j] == fav[i])
				dp[i][j] = maxn + 1;
			else
				dp[i][j] = maxn;
		}
	}
//	for(int i = 1; i <= m; ++i){
//		for(int j = 1; j <= l; ++j){
//			printf("%d ",dp[i][j]);
//		}
//		cout <<endl; 
//	}
	
	printf("%d",dp[m][l]);
	return 0;
}

猜你喜欢

转载自blog.csdn.net/w419387229/article/details/82143505
今日推荐