uva10453 Make Palindrome（DP 求方案）

if(s[x]==s[y]) dp[x][y]=dp[x+1][y-1]+2

else dp[x][y]=min(dp[x+1][y],dp[x][y-1])+2

方案分开来求

#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <algorithm>
using namespace std;
char s[1010];
int dp[1010][1010];
int x,y,ac,l;
string ans;
int DP(){
    if(dp[x][y]!=-1) return dp[x][y];
    if(x>y) return dp[x][y]=0;
    if(x==y) return dp[x][y]=1;
    int ret=10000;
    if(s[x]==s[y]){
        x++,y--;
        ret=min(ret,DP()+2);
        x--,y++;
    }else{
        x++;
        ret=min(ret,DP()+2);
        x--;
        y--;
        ret=min(ret,DP()+2);
        y++;
    }
    //cout<<x<<" "<<y<<" "<<ret<<endl;
    return dp[x][y]=ret;
}
void gao(){
    if(x>y) return;
    if(x==y) {ans+=s[x];return;}
    //cout<<x<<" "<<y<<" "<<ac<<endl;
    if(s[x]==s[y]){
        ac-=2;
        ans+=s[x];
        x++,y--;
        gao();
    }else if(dp[x][y-1]+2==ac&&dp[x][y]==dp[x][y-1]+2){
        ac-=2;
        ans+=s[y];
        y--;
        gao();
    }else if(dp[x+1][y]+2==ac&&dp[x][y]==dp[x+1][y]+2){
        ac-=2;
        ans+=s[x];
        x++;
        gao();
    }
}
void solve(){
    l=strlen(s+1);
    memset(dp,-1,sizeof dp);
    x=1,y=l;
    int ret=DP();
    printf("%d ",ret-l);
    ac=ret;
    ans="";
    x=1,y=l;
    gao();
    int len=ans.length();
    for(int i=0;i<len;i++) putchar(ans[i]);
    int start=ret&1 ? len-2:len-1;
    for(int i=start;i>=0;i--) putchar(ans[i]);
    putchar('\n');
}
int main(){
    //freopen("10453.txt","r",stdin);
    while(~scanf("%s",s+1)){
        solve();
    }
    return 0;
}