poj 1159

这题用short水过，一看网上大牛的代码，居然说可以用滚动数组！太厉害了。

//poj_1159
#include <stdio.h>
#include <string.h>
#define N	5010
#define INF	10000
#define hpmin(a,b)	(a)<(b)?(a):(b)
int n;
char str[N];
short int dp[N][N];
int main()
{
	int i, len, j;
	while ( scanf("%d", &n ) != EOF ) {
		scanf("%s", str );
		for ( i = 0; i < n; ++ i ) {
			dp[i][i] = 0;
			dp[i+1][i] = 0;
		}
		for ( len = 1; len < n; ++ len ) {	//我总喜欢先枚举长度，再枚举起点i，计算终点j。如果是这样子，能难使用滚动数组	
			for ( i = 0; i < n-len; ++ i ) {
				j = i+len;
				dp[i][j] = INF;
				if ( str[i] == str[j] ) {
					dp[i][j] = dp[i+1][j-1];
				}
				dp[i][j] = hpmin( dp[i][j], dp[i+1][j]+1 );
				dp[i][j] = hpmin( dp[i][j], dp[i][j-1]+1 );
			}
		}
		printf("%d\n", dp[0][n-1] );
	}
	return 0;
}

滚动数组版本：

//poj_1159
#include <stdio.h>
#include <string.h>
#define N	5010
#define INF	10000
int hpmin( int x, int y ) {
	return ( x < y ? x : y );
}
int n;
char str[N];
short int dp[2][N];
int main()
{
	int i, j, a, b;
	while ( scanf("%d", &n ) != EOF ) {
		scanf("%s", str );
		memset( dp, 0, sizeof(dp) );
		a = 0;
		for ( i = n-2; i >= 0; -- i ) {	//直接先枚举i，再枚举j就很容易看出使用滚动数组。dp（i,j) = max( dp(i+1,j),dp(i,j-1), dp(i+1,j-1) )这种形式的dp应该都可以使用滚动数组优化
			b = 1-a ;
			dp[b][i] = 0;
			for ( j = i+1; j < n; ++ j ) {
				dp[b][j] = INF;
				if ( str[i] == str[j] )
					dp[b][j] = dp[a][j-1];
				//dp[b][j] = hpmin( dp[b][j], dp[a][j]+1 );
				//dp[b][j] = hpmin( dp[b][j], dp[b][j-1]+1 );
				else {
					//int tmp = hpmin( dp[a][j], dp[b][j-1] );
					dp[b][j] = hpmin( dp[a][j], dp[b][j-1] ) + 1;
				}
			}
			a = b;
			//for ( j = 0; j < n; ++ j )
			//	printf("%d ", dp[b][j] );
			//printf("\n");
		}
		printf("%d\n", dp[a][n-1] );
	}
	return 0;
}