• UVA 10651 Pebble Solitaire 状态压缩dp


    一开始还在纠结怎么表示一个状态,毕竟是一个串。后来搜了一下题解发现了这里用一个整数的前12位表示转态就好了 ,1~o,0~'-',每个状态用一个数来表示,然后dp写起来就比较方便了。

    代码:

     1 #include <iostream>
     2 #include <sstream>
     3 #include <cstdio>
     4 #include <climits>
     5 #include <cstring>
     6 #include <cstdlib>
     7 #include <string>
     8 #include <cmath>
     9 #include <vector>
    10 #include <queue>
    11 #include <algorithm>
    12 #define esp 1e-6
    13 #define pb push_back
    14 #define in  freopen("in.txt", "r", stdin);
    15 #define out freopen("out.txt", "w", stdout);
    16 #define print(a) printf("%d
    ",(a));
    17 #define bug puts("********))))))");
    18 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
    19 #define inf 0x0f0f0f0f
    20 using namespace std;
    21 typedef long long  LL;
    22 typedef vector<int> VI;
    23 typedef pair<int, int> pii;
    24 typedef vector<pii,int> VII;
    25 typedef vector<int>:: iterator IT;
    26 #define N 50000
    27 int dp[N];
    28 int ans;
    29 void f(int x)
    30 {
    31     if(dp[x])
    32         return ;
    33     int num = 0;
    34     for(int i = 0; i < 12; i++)
    35         if(x&(1<<i))
    36         num++;
    37     ans = min(ans, num);
    38     dp[x] = 1;
    39     for(int i = 0; i <= 9; i++)
    40         if(((x&(1<<i)) && (x&(1<<(i+1))) && !(x&(1<<(i+2))))
    41                 ||(!(x&(1<<i)) && (x&(1<<(i+1))) && (x&(1<<(i+2)))))
    42             f(x^(1<<(i+1))^(1<<i)^(1<<(i+2)));
    43 }
    44 int main(void)
    45 {
    46 
    47     char s[100];
    48     int T, t;
    49     for(t = scanf("%d", &T), gets(s); t <= T; t++)
    50     {
    51         memset(dp, 0, sizeof(dp));
    52         gets(s);
    53         int n = 0;
    54         ans = 20;
    55         for(int i = 0; i < 12; i++)
    56             if(s[i] - '-')
    57                 n ^= (1<<i);
    58                 f(n);
    59         printf("%d
    ", ans);
    60     }
    61     return 0;
    62 }
    View Code

    或者可以用map+string的方法,一直都没怎么学过STL,看了http://www.myexception.cn/ai/1243266.html 这里的方法表示又涨了不少知识,拿来重新写了一遍(其实还是“剽窃”,没办法先当一个"搬运工”吧)。

     1 #include <iostream>
     2 #include <sstream>
     3 #include <cstdio>
     4 #include <climits>
     5 #include <cstring>
     6 #include <cstdlib>
     7 #include <string>
     8 #include <cmath>
     9 #include <stack>
    10 #include <map>
    11 #include <vector>
    12 #include <queue>
    13 #include <algorithm>
    14 #define esp 1e-6
    15 #define pb push_back
    16 #define in  freopen("in.txt", "r", stdin);
    17 #define out freopen("out.txt", "w", stdout);
    18 #define print(a) printf("%d
    ",(a));
    19 #define bug puts("********))))))");
    20 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
    21 #define inf 0x0f0f0f0f
    22 using namespace std;
    23 typedef long long  LL;
    24 typedef vector<int> VI;
    25 typedef pair<int, int> pii;
    26 typedef vector<pii,int> VII;
    27 typedef vector<int>:: iterator IT;
    28 map<string, bool> my;
    29 int ans;
    30 void dfs(string cur)
    31 {
    32     if(my.find(cur) != my.end())
    33         return;
    34     int len = cur.size(), num = 0;
    35     for(int i = 0; i < len; i++)
    36         if(cur[i] == 'o')
    37         num++;
    38     ans = min(ans, num);
    39     my[cur] = true;
    40     for(int i = 0; i <= 9; i++)
    41     {
    42         if(cur.substr(i, 3) == "-oo")
    43         {
    44             string temp = cur;
    45             temp.replace(i, 3, "o--");
    46             dfs(temp);
    47         }
    48         if(cur.substr(i, 3) == "oo-")
    49         {
    50             string temp = cur;
    51             temp.replace(i, 3, "--o");
    52             dfs(temp);
    53         }
    54     }
    55 }
    56 int main(void)
    57 {
    58     int T;
    59     string s;
    60     for(int t = scanf("%d", &T); t <= T; t++)
    61     {
    62         cin>>s;
    63         my.clear();
    64         ans = 20;
    65         dfs(s);
    66         printf("%d
    ", ans);
    67     }
    68     return 0;
    69 }
    View Code
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  • 原文地址:https://www.cnblogs.com/rootial/p/3329342.html
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