URAL1635——DP+回溯——Mnemonics and Palindromes

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char  a[4200];
int dp[4200];
int b[4200][4200];
int path[4200];
int n;
char pri[4200][4200];
void inti(){
    for(int i = 1 ; i <= n; i++){
        for(int j = i + 1 ;j <= n; j++){
            int k,p;
            for( k = i,p = j; k < p; k++,p--){
                if(a[k] != a[p] )
                        break;
            }
            if(k >= p)
            b[i][j] = j - i + 1;
        }
    }
    for(int i = 1 ; i <= n; i++)
        b[i][i] = 1;

}
int main()
{
    while(~scanf("%s",a+1)){
        memset(b,0,sizeof(b));
        memset(dp,0,sizeof(dp));
        memset(path,0,sizeof(path));
     n = strlen(a+1);
    inti();
    dp[1] = 1;
    for(int i = 2 ; i <= n;i++)
        dp[i] = dp[i-1]+1;
    dp[0] = 0;
    for(int i = 1; i <= n; i++){
        for(int j = i ; j <= n; j++){
            if(dp[j-b[i][j]]+1 <= dp[j]){
                dp[j] = dp[j-b[i][j]] + 1;
                path[j] = j - b[i][j];
            }
        }
    }
    //for(int i = 1; i <= n ;i++)
    //printf("%d\n",path[i]);
     // for(int i = 1; i <= n; i++)
     //   for(int j = i+1 ; j <= n;j++)
     //       printf("%d ",b[i][j]);
     //  for(int i = 1 ; i <= n; i++)
     //      printf("%d ",dp[i]);
    printf("%d\n",dp[n]);
    int i = n ;
    int t1 = 1;
    int t2;
    memset(pri,0,sizeof(pri));
    while( i >= 1){  
         t2 = 1;
      for(int j = path[i]+1; j <= i ;j++){
          pri[t1][t2++] = a[j];
        // printf("%d%d%c ",t1,t2-1,pri[t1][t2-1]);
      }
    //  printf("\n");
      t1++;
      i = path[i];
    }
 //   printf("%d",t1);
   for(int i = t1 - 1; i >= 1; i--){
        for(int j = 1; pri[i][j] != ‘\0‘;j++)
           printf("%c",pri[i][j]);
       if(i!=1) printf(" ");
    }
   printf("\n");
    }
    return 0;
}

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[4010];
int pre[4010];
int ok[4010][4010];
char s[4010];
void print(int p)
{
    if(pre[p] == -1){
        for(int i = 0 ; i <= p ;i++){
            printf("%c",s[i]);
        }
    }
       else{
        print(pre[p]);
        printf(" ");
        for(int i = pre[p]+1; i <= p ;i++)
            printf("%c",s[i]);
    }
}

void solve()
{
    int len = strlen(s);
    for(int i = 0;i < len ;i++){//i表示长度
        for(int j = 0 ;i+j < len;j++){
            if(i == 0) ok[j][j+i] = 1; //因为从小到大所以是符合这个方程的   
            else if(i == 1) ok[j][j+i] = (s[j] == s[j+i]);
            else if(s[j] == s[j+i])
                ok[j][j+i] = ok[j+1][j+i-1];
        }
    }
    for(int i = 0; i < len ;i++){
        dp[i] = i + 1;
        pre[i] = i - 1;//从0到i是否回文
        if(ok[0][i]) {
            dp[i] = 1;
            pre[i] = -1;
            continue;
        }
        for(int j = i - 1; j >= 0 ; j--){
            if(ok[j+1][i]){//j表示i前面
                if(dp[i] > dp[j] + 1){
                    dp[i] = dp[j] + 1;
                    pre[i] = j;
                }
            }
        }
    }
    printf("%d\n",dp[len-1]);
    print(len-1);//dfs
    printf("\n");
}

int main()
{
    while(~scanf("%s",s)){
        memset(dp,0,sizeof(dp));
        memset(pre,0,sizeof(pre));
        memset(ok,0,sizeof(ok));
        solve();
    }
    return 0;
}