POJ 1159

区间dp

比如从i到j如果str[i] == str[j]那么i到j的值就与i+1到j-1相同,否则从i到j-1与i+1到j中选一个最优解,这道题直接定义int的二维数组会mle,用short,也可以用滚动数组。

#include<stdio.h>
const int maxa = 5005;
short dp[maxa][maxa];
char str[maxa];
#define min(a,b) a<b?a:b;
int main(){
    int n;
    while(scanf("%d", &n)!=EOF){
        scanf("%s", str);
        for(int i = 1; i < n; i++){
            for(int 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] = min(dp[j+1][j+i]+1, dp[j][j+i-1]+1);
                }
            }
        }
        printf("%d
", dp[0][n-1]);
    }
}
View Code
原文地址:https://www.cnblogs.com/icodefive/p/4299684.html