UVA 531 - Compromise

最长公共子序列问题，打印路径总结了两个方法，一个是递归指向，就像前面做的01背包问题如：

for(int i = 1; i < n; i ++)
        for(int j = 0; j < n-i; j ++)
            for(int k = j; k < i+j; k ++)
            {
                if(k == j) 
                {
                    f[j][i+j] = f[j][k]+f[k+1][i+j]+A[j][0]*A[k][1]*A[i+j][1];
                    part1[j][i+j] = k;//赋值
                }
                if(f[j][k]+f[k+1][i+j]+A[j][0]*A[k][1]*A[j+i][1] < f[j][i+j]) 
                    {
                        f[j][i+j] = f[j][k]+f[k+1][i+j]+A[j][0]*A[k][1]*A[j+i][1];
                        part1[j][i+j] = k;//赋值
                    }
            }

void printpath(int a, int b)//输出
{
    if(a == b)
    {
        printf("A%d", a + 1);
        return ;
    }
    printf("(");
    printpath(a, part1[a][b]);
    printf(" x ");
    printpath(part1[a][b] + 1, b);
    printf(")");
}

例如：

for(int i = 1; i <= cdnum; i ++)
        for(int j = 0; j <= tap; j ++)
        {
            f[i][j] = f[i-1][j];
            p[i][j] = j;//赋值
            if(j>=num[i]&&f[i-1][j-num[i]]+num[i] > f[i][j])
            {
                f[i][j] = f[i-1][j-num[i]]+num[i]; p[i][j] = j-num[i];//赋值
            }
        }

void print(int i, int j)//打印路径
{
    if(i == 0)
        return;
    print(i - 1, p[i][j]);
    if(p[i][j] < j)
        printf("%d ", num[i]);
}

这第二种就是今天用到的 赋值遍历打印路径；

    for(int i = 1; i <= m; i ++)
        for(int j = 1; j <= n; j ++)
            if(!strcmp(a[i], b[j])) 
                {
                    f[i][j] = f[i-1][j-1] + 1;
                    p[i][j] = 1;//赋值
                }
            else if(f[i-1][j]>f[i][j-1])
                        {
                            f[i][j] = f[i-1][j];
                            p[i][j] = 0;//赋值
                        }
                        else 
                        {
                            f[i][j] = f[i][j-1];
                            p[i][j] = -1;//赋值
                        }

void print(int i , int j)//输出
{
    if(!i || !j)
        return ;
    if(p[i][j] == 1)
    {
        print(i-1,j-1);
        if(flag)
      printf(" ");
    else
      flag = 1;
    printf("%s", a[i]);
    }
    else if(p[i][j] == 0)
                print(i-1,j);
                else print(i,j-1);
}

代码如下：

#include<stdio.h>
#include<string.h>
#define MAXN 100 + 5
char a[MAXN][35], b[MAXN][35], f[MAXN][MAXN], p[MAXN][MAXN];
int m, n, flag;
void print(int i , int j)
{
    if(!i || !j)
        return ;
    if(p[i][j] == 1)
    {
        print(i-1,j-1);
        if(flag)
      printf(" ");
    else
      flag = 1;
    printf("%s", a[i]);
    }
    else if(p[i][j] == 0)
                print(i-1,j);
                else print(i,j-1);
}
void solve()
{
    memset(f,0,sizeof(f));
    for(int i = 1; i <= m; i ++)
        for(int j = 1; j <= n; j ++)
            if(!strcmp(a[i], b[j])) 
                {
                    f[i][j] = f[i-1][j-1] + 1;
                    p[i][j] = 1;
                }
            else if(f[i-1][j]>f[i][j-1])
                        {
                            f[i][j] = f[i-1][j];
                            p[i][j] = 0;
                        }
                        else 
                        {
                            f[i][j] = f[i][j-1];
                            p[i][j] = -1;
                        }
    print(m,n);
    printf("\n");
}
void input()
{
    while(scanf("%s",a[1]) == 1)
    {
        flag = 0;
        m = 1;
        n = 0;
        if(a[1][0] != '#')
        for(int j = 2; ;j ++)
        {
            scanf("%s",a[j]);
            if(a[j][0] == '#') break;
            m ++;
        }
        for(int i = 1; ; i ++)
        {
            scanf("%s",b[i]);
            if(b[i][0] == '#') break;
            n ++;
        }
        solve();
    }
}
int main()
{
    input();
    return 0;    
}