uva_10651_Pebble Solitaire(状态压缩DP)

题目意思很明确，12这个数字很有状态压缩的意味这一题的阶段就是按照left -> right 的顺序DP这里一共有2^12个状态，从1 到 2^12 的数字，每个数字的二进制表示这一行的状态状态:dp[i]表示该状态的游戏最终结果状态转移：dp[i] = min(dp[j]）, j就是从i的状态经过游戏规则所能转换的状态ans = dp[n]#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define MAXBIT  12#define INF     0x3f3f3f3f#define MAXN    1<<(MAXBIT+1)int dp[MAXN];int dfs(const int &idx){        if( -1 != dp[idx] ) {                return dp[idx];        }        int val(idx), rst(INF), flag(0), cnt(0), tmp;        for(int i = 0; i < MAXBIT; i ++) {                cnt += ((1<<i)&val)? 1 : 0;                if( (1<<i)&val && i+1 < MAXBIT && (1<<(i+1))&val ) {                        if( i > 0 && !((1<<(i-1))&val) ) {              //move to left                                tmp = val; tmp ^= (1<<(i-1));                                tmp ^= (1<<(i+1)); tmp ^= (1<<i); flag = 1;                                rst = min(rst, dp[tmp] = (-1 == dp[tmp])? dfs(tmp) : dp[tmp]);                        }                        if( i+2 < MAXBIT && !((1<<(i+2))&val) ) {       //move to right                                tmp = val; tmp ^= (1<<(i+2));                                tmp ^= (1<<(i+1)); tmp ^= (1<<i); flag = 1;                                rst = min(rst, dp[tmp] = (-1 == dp[tmp])? dfs(tmp) : dp[tmp]);                        }                }        }        if( !flag ) {                return dp[idx] = cnt;        }        return dp[idx] = rst;}int conversion(const char *str){        int rst(0);        for(int i = strlen(str)-1; i >= 0; i --) {                if( 'o' == str[i] ) {                        rst += (1<<(strlen(str)-i-1));                }        }        return rst;}int main(int argc, char const *argv[]){#ifndef ONLINE_JUDGE        freopen("test.in", "r", stdin);#endif        int n;        char str[MAXBIT+1];        memset(dp, -1, sizeof(dp));        for(int i = 1<<MAXBIT; i >= 0; i --) {                dp[i] = (-1 == dp[i])? dfs(i) : dp[i];        }        while( ~scanf("%d", &n) ) {                for(int i = 0; i < n; i ++) {                        scanf("%s", str);                        printf("%d\n", dp[ conversion(str) ]);                }        }        return 0;}