UVA - 11404

题意：求任意删除字符后所形成的最长回文，并输出字典序最小的方案

把原串反转求LIS，因为转移时不断求字典序最小导致后半部分可能并非回文，所以要前半部分输出两边

话说这方案保存可真暴力

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cmath>

#include<string>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#define rep(i,j,k) for(register int i=j;i<=k;i++)

#define rrep(i,j,k) for(register int i=j;i>=k;i--)

#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])

#define iin(a) scanf("%d",&a)

#define lin(a) scanf("%lld",&a)

#define din(a) scanf("%lf",&a)

#define s0(a) scanf("%s",a)

#define s1(a) scanf("%s",a+1)

#define print(a) printf("%lld",(ll)a)

#define enter putchar('\n')

#define blank putchar(' ')

#define println(a) printf("%lld\n",(ll)a)

#define IOS ios::sync_with_stdio(0)

using namespace std;

const int maxn = 1e3+11;

const double eps = 1e-10;

typedef long long ll;

const int oo = 0x3f3f3f3f;

ll read(){

    ll x=0,f=1;register char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

char str[maxn],strr[maxn];

int dp[maxn][maxn];

string ans[maxn][maxn];

int main(){

    while(~s1(str)){

        int n=strlen(str+1);

        rep(i,1,n) strr[n-i+1]=str[i];

        memset(dp,0,sizeof dp);

        rep(i,0,n)rep(j,0,n) ans[i][j]=" ";

        int len=0;

        rep(i,1,n) rep(j,1,n){

            if(str[i]==strr[j]){

                dp[i][j]=dp[i-1][j-1]+1;

                ans[i][j]=ans[i-1][j-1]+str[i];

            }else{

                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);

                if(dp[i-1][j]>dp[i][j-1]) ans[i][j]=ans[i-1][j];

                else if(dp[i][j-1]>dp[i-1][j]) ans[i][j]=ans[i][j-1];

                else if(ans[i][j-1]<ans[i-1][j]) ans[i][j]=ans[i][j-1];

                else ans[i][j]=ans[i-1][j];

            }

        }

        if(dp[n][n]&1){

            rep(i,1,(dp[n][n]+1)/2) cout<<ans[n][n][i];

            rep(i,1,dp[n][n]/2) cout<<ans[n][n][dp[n][n]/2-i+1];

            cout<<endl;

        }else{

            rep(i,1,dp[n][n]/2) cout<<ans[n][n][i];

            rep(i,1,dp[n][n]/2) cout<<ans[n][n][dp[n][n]/2-i+1];

            cout<<endl;

        }

        // cout<<ans[n][n]<<endl;

    }

    return 0;

}