UVA 10453 Make Palindrome

UVA_10453

    这个题目要求打印最后的结果，所以我们在动规的过程中要记录一下当前的决策，便于我们后面递归去打印回文串。

#include<stdio.h>
#include<string.h>
#define MAXD 1010
#define INF 1000000000
int N, f[MAXD][MAXD], p[MAXD][MAXD];
char b[MAXD];
void printpath(int x, int y)
{
    if(x > y)
        return;
    if(p[x][y] == 0)
    {
        if(x == y)
            printf("%c", b[x]);
        else
        {
            printf("%c", b[x]);
            printpath(x + 1, y - 1);
            printf("%c", b[y]);
        }
    }
    else if(p[x][y] == 1)
    {
        printf("%c", b[y]);
        printpath(x, y - 1);
        printf("%c", b[y]);
    }
    else
    {
        printf("%c", b[x]);
        printpath(x + 1, y);
        printf("%c", b[x]);
    }
}
void solve()
{
    int i, j, k;
    N = strlen(b + 1);
    for(i = 0; i <= N; i ++)
        for(j = i + 1; j <= N; j ++)
            f[i][j] = INF;
    for(i = 1; i <= N; i ++)
        f[i][i - 1] = f[i][i] = 0;
    for(i = 1; i <= N; i ++)
        p[i][i] = 0;
    for(k = 1; k <= N; k ++)
        for(i = 1; i + k <= N; i ++)
        {
            if(b[i] == b[i + k])
            {
                p[i][i + k] = 0;
                f[i][i + k] = f[i + 1][i + k - 1];
            }
            else
            {
                if(f[i][i + k - 1] + 1 < f[i][i + k])
                {
                    f[i][i + k] = f[i][i + k - 1] + 1;
                    p[i][i + k] = 1;
                }
                if(f[i + 1][i + k] + 1 < f[i][i + k])
                {
                    f[i][i + k] = f[i + 1][i + k] + 1;
                    p[i][i + k] = 2;
                }
            }
        }
    printf("%d ", f[1][N]);
    printpath(1, N);
    printf("\n");
}
int main()
{
    while(scanf("%s", b + 1) == 1)
    {
        solve();
    }
    return 0;
}