【DP】UVA 10651 Pebble Solitaire 记忆化搜索

来源：互联网发布：视频解码软件编辑：程序博客网时间：2024/05/18 00:25
#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#include <string>#include <iostream>#include <algorithm>using namespace std;#include <queue>#include <stack>#include <vector>#include <deque>#include <set>#include <map>#define cler(arr, val)    memset(arr, val, sizeof(arr))#define IN     freopen ("in.txt" , "r" , stdin);#define OUT  freopen ("out.txt" , "w" , stdout);typedef long long  LL;const int MAXN = 111;//点数的最大值const int MAXM = 20006;//边数的最大值const int INF = 11521204;const int mod=1000000007;int dp[1<<14];void print(int x){    int out=0;    while(x)    {        int a=x%2;        if(a) out++;        x/=2;    }    printf("%d\n",out);}int bfs(int x){    queue<int>q;    q.push(x);    while(!q.empty())    {        x=q.front();        q.pop();        for(int i=0; i<12; i++)        {            if((1 << i) & x)            {                if(i<10&&(x&(1<<(i+1)))&&!(x&(1<<(i+2))))                {                    int ans=x^(1<<(i+1));                    ans=ans^(1<<(i+2));                    ans=ans^(1<<i);                    if(!dp[ans])                    {                        dp[ans]=1;                        q.push(ans);                    }                }                if(i>2&&(x&(1<<(i-1)))&&!(x&(1<<(i-2))))                {                    int ans=x^(1<<(i-1));                    ans=ans^(1<<(i-2));                    ans=ans^(1<<i);                    if(!dp[ans])                    {                        dp[ans]=1;                        q.push(ans);                    }                }            }        }    }    print(x);}int main(){    char s[13];    int t;  //  IN;    scanf("%d",&t);    while(t--)    {        cler(dp,0);        scanf("%s",s);        int in=0,x=1;        for(int i=0; i<12; i++)        {            in+=x*(s[i]=='o'?1:0);            x*=2;        }        bfs(in);    }    return 0;}
0 0