acdream 1025 简单dp

最基本的在DAG上求最短路。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int INF = 999999;
const int N = 100001;
int dp[N];

int main ()
{
    int a, b;
    while ( scanf("%d%d", &a, &b) != EOF )
    {
        if ( a > b )
        {
            printf("-1
");
            continue;
        }
        dp[a] = 0;
        for ( int i = a + 1; i <= b; i++ )
        {
            dp[i] = INF;
        }    
        for ( int i = a; i < b; i++ )
        {
            int j;
            for ( j = 1; j * j < i; j++ )
            {
                if ( i % j ) continue;
                int x = i + j;
                if ( x <= b )
                {
                    dp[x] = min( dp[x], dp[i] + 1 );
                }
                int y = i + i / j;
                if ( y <= b )
                {
                    dp[y] = min( dp[y], dp[i] + 1 );
                }
            }
            if ( i % j == 0 )
            {
                int z = i + j;
                if ( z <= b )
                {
                    dp[z] = min( dp[z], dp[i] + 1 );
                }
            }
        }
        printf("%d
", dp[b]);
    }
    return 0;
}