poj3280 构造回文字符串最小花费-区间dp

题意

思路

dp[i][j]保存的是i~j这一段已经是回文且最优解

那么当我们要求新的区段i~j时只需要考虑s[i] ?= s[j]

if(s[i] == s[j]) dp[i][j] = min(dp[i][j],dp[i +1][j - 1]);

else dp[i][j] = min(dp[[i][j],min(dp[i + 1][j] + chan[s[i]],dp[i][j - 1] + chan[s[j]]));

chan数组每个字符min(add,reduce)，这是因为当两边s[i] != s[j]时 对于从dp[i +1][j]转化过来 s[i]可以在i这一位舍弃或者s[j]后面补上一个s[i] 因此cost += min(add,reduce) dp[i][j - 1]也是一样的过程

每个dp[i][j]只代表这一段的回文最优解 是无后效性的 直接继承就好不会影响后面的答案

ACcode

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>

using namespace std;
const int inf = 0x3f3f3f3f;

int n,len;
char op;
char s[2005];
int ad[30],re[30],chan[30];
int dp[2005][2005];

int main()
{
	scanf("%d %d",&n,&len);
	scanf("%s",s + 1);
	s[0] = '#',s[len + 1] = '#';
	for(int i = 1;i <= n; i++){
		scanf(" %c",&op);
		int cur = op - 'a';
		scanf("%d %d",&ad[cur],&re[cur]);
		chan[cur] = min(ad[cur],re[cur]);
	}
	memset(dp,0,sizeof dp);
	for(int i = 1;i <= len; i++){
		for(int j = i;j <= len; j++) dp[i][j] = inf;
	}
	for(int L = 1;L <= len; L++){
		for(int i = 1;i + L - 1 <= len; i++){
			int j = i + L - 1;
			if(s[i] == s[j]) dp[i][j] = min(dp[i][j],dp[i + 1][j - 1]);
			else dp[i][j] = min(dp[i][j],min(dp[i + 1][j] + chan[s[i] - 'a'],dp[i][j - 1] + chan[s[j] - 'a']));
		}
	}
	printf("%d\n",dp[1][len]);
}