UVa10651(记忆化搜索)

题意:

给一串-和o组成的字符串,你可以把“-oo"变成”o--",可以把“oo-”变成“--o",问最后最少有多少个o.


状态压缩 ,记忆化搜索

code:

#include <stdio.h>
#include <string.h>
#define len 12
#define min(a,b) (((a)<(b)) ?(a):(b))
int d[1<<13];
int dp(int n)
{
    int i, t;
    if(d[n]!=-1) return d[n];
    d[n] = 0;
    for(i=0; i<len; i++)
        if(n&(1<<i))
            d[n]++;
    for(i=0; i<len-2; i++)
    {
        t = n;
        if( (t&(1<<i)) && (t&(1<<(i+1))) && !(t&(1<<(i+2))) )
        {
            t &=~(1<<i);
            t &=~(1<<(i+1));
            t |=1<<(i+2);
            d[n] = min(d[n],dp(t));
        }
        if( !(t&(1<<i)) && (t&(1<<(i+1))) && (t&(1<<(i+2))) )
        {
            t &=~(1<<(i+1));
            t &=~(1<<(i+2));
            t |=1<<i;
            d[n] = min(d[n],dp(t));
        }
    }
    return d[n];
}
int main()
{
    int T, i, n;
    char str[20];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",str);
        n = 0;
        for(i=0; i<len; i++)
            if(str[i]=='o')
                n ^=1<<i;
        memset(d,-1,sizeof(d));
        printf("%d
", dp(n));
    }
    return 0;
}



原文地址:https://www.cnblogs.com/javawebsoa/p/3217928.html