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两个代码,用的都是DP,只是运算顺序不同,结果一个AC一个TLE

Posted by linandmin at 2006-07-29 23:25:05 on Problem 1159
这是AC的代码:

#include<stdio.h>
#include<iostream>
using namespace std;

char str[5010];
short dp[5010][5010]={0};

short minimum(short a,short b)
{
	return a<b?a:b;
}

int main()
{
	int n,i,j;
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%s",str+1);
		for(i=n;i>0;i--)
			for(j=0;j<=n;j++)
			{
				if(i>=j) continue;
				if(str[i]==str[j])
					dp[i][j]=dp[i+1][j-1];
				else
					dp[i][j]=minimum(dp[i+1][j],dp[i][j-1])+1;
			}
		printf("%d\n",dp[1][n]);
	}
	return 0;
}

这是TLE的代码:
#include<stdio.h>
#include<iostream>
using namespace std;

char str[5010];
short dp[5010][5010]={0};

short minimum(short a,short b)
{
	return a<b?a:b;
}

int main()
{
	int n,i,j;
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%s",str);
		for(i=1;i<=n-1;i++)
			for(j=0;j+i<n;j++)
			{
				if(str[j]==str[j+i])
					dp[j][j+i]=dp[j+1][j+i-1];
				else
					dp[j][j+i]=minimum(dp[j+1][j+i],dp[j][j+i-1])+1;
			}
		printf("%d\n",dp[0][n-1]);
	}
	return 0;
}

这两代码用的算法都是DP,DP状态也一样,dp[i][j]为第i个字符和第j个字符之间所要插入的最少的字符个数。转移方程为dp[i][j]=dp[i+1][j-1](str[i]==str[j]),dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1;(str[i]!=str[j])。

而且AC的那份代码过不了这组数据:
5000
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

哪位大哥能解释一下吗?

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