简单DP（51nod 1092）

Posted 2020-07-04 yoyo_sincerely

题目：回文字符串

思路：找准状态以及决策，就可以了；

 形如：E[i,j]=opt{D[i-1,j]+xi,D[i,j-1]+yj,D[i-1][j-1]+zij} (最长公共子序列)   

　变形即可；

dp[i][j]是第i位置开始长度为j的最小添加的字符串的数量；

 

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <time.h>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>

#define c_false ios_base::sync_with_stdio(false); cin.tie(0)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
#define swa(x,y) {LL s;s=x;x=y;y=s;}
using namespace std ;
#define N 1005

const double PI = acos(-1.0);
typedef long long LL ;
int dp[N][N];
char a[N];
int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    //ios_base::sync_with_stdio(false); cin.tie(0);
    scanf("%s", a);
    int n = strlen(a);
    memset(dp,0,sizeof(dp));
    for(int j = 1; j <= n; j++){
        for(int i = 0; j+i-1 < n; i++){
            dp[i][j] = min(dp[i][j-1], dp[i+1][j-1]) +1;
            if(a[i] == a[i+j-1]) dp[i][j] = min(dp[i+1][j-2], dp[i][j]);
        }
    }
    cout<<dp[0][n]<<endl;
    return 0;
}