POJ3185(簡單BFS，主要做測試使用)

沒事做水了一道POJ的簡單BFS的題目

這道題的數據范圍是20,所以狀態總數就是（1<<20）

第一次提交使用STL的queue，并且是在隊首判斷是否達到終點，達到終點就退出，超時：（其實這里我是很不明白的，，TM狀態總數就只有1e6怎么也不應該超時的，，，，只能說STL的queue的常數實在是太大，完全沒法弄。。。）

 1 #include <map>
 2 #include <set>
 3 #include <stack>
 4 #include <queue>
 5 #include <cmath>
 6 #include <ctime>
 7 #include <vector>
 8 #include <cstdio>
 9 #include <cctype>
10 #include <cstring>
11 #include <string.h>
12 #include <cstdlib>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16 #define INF 1e9
17 #define inf (-((LL)1<<40))
18 #define lson k<<1, L, mid
19 #define rson k<<1|1, mid+1, R
20 #define mem0(a) memset(a,0,sizeof(a))
21 #define mem1(a) memset(a,-1,sizeof(a))
22 #define mem(a, b) memset(a, b, sizeof(a))
23 #define FOPENIN(IN) freopen(IN, "r", stdin)
24 #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
25 template<class T> T CMP_MIN(T a, T b) { return a < b; }
26 template<class T> T CMP_MAX(T a, T b) { return a > b; }
27 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
28 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
29 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
30 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;      }
31 template<class T> void SWAP(T& a, T& b) { T x = a; a = b; b = x; }
32 
33 //typedef __int64 LL;
34 typedef long long LL;
35 const int MAXN = 200010;
36 const int MAXM = 100005;
37 const double eps = 1e-10;
38 //const LL MOD = 1000000007;
39 
40 int step[1<<21];
41 bool vis[1<<21];
42 
43 int BFS(int s)
44 {
45     vis[s] = 1;
46     queue<int>q;
47     q.push(s);
48     step[s] = 0;
49     while(!q.empty())
50     {
51         int u = q.front(); q.pop();
52         if(u == 0)  return step[u];
53         for(int i=1;i<19;i++)
54         {
55             int r = u;
56             r ^= (1<<i-1) | (1<<i) | (1<<i+1);
57             if(!vis[r])
58             {
59                 vis[r] = 1;
60                 step[r] = step[u] + 1;
61                 q.push(r);
62             }
63         }
64         int r = u ^ (1<<0) ^ (1<<1);
65         if(!vis[r]) { vis[r] = 1; step[r] = step[u] + 1; q.push(r); }
66         r = u ^ (1<<18) ^ (1<<19);
67         if(!vis[r]) { vis[r] = 1; step[r] = step[u] + 1; q.push(r); }
68     }
69     return -1;
70 }
71 
72 int main()
73 {
74         //FOPENIN("in.txt");
75         int st = 0, x;
76         for(int i=0;i<20;i++)
77         {
78             scanf("%d", &x);
79             st |= (x<<i);
80         }
81         printf("%d\n", BFS(st));
82         return 0;
83 }

View Code

TLE后馬上把判斷放到隊尾（就是說在一個狀態進隊列前先判斷是不是終點狀態，是的話就退出），跑了875ms，勉強過了，，（這里我就是亂改的了，代碼沒任何觀賞性）

 1 #include <map>
 2 #include <set>
 3 #include <stack>
 4 #include <queue>
 5 #include <cmath>
 6 #include <ctime>
 7 #include <vector>
 8 #include <cstdio>
 9 #include <cctype>
10 #include <cstring>
11 #include <string.h>
12 #include <cstdlib>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16 #define INF 1e9
17 #define inf (-((LL)1<<40))
18 #define lson k<<1, L, mid
19 #define rson k<<1|1, mid+1, R
20 #define mem0(a) memset(a,0,sizeof(a))
21 #define mem1(a) memset(a,-1,sizeof(a))
22 #define mem(a, b) memset(a, b, sizeof(a))
23 #define FOPENIN(IN) freopen(IN, "r", stdin)
24 #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
25 template<class T> T CMP_MIN(T a, T b) { return a < b; }
26 template<class T> T CMP_MAX(T a, T b) { return a > b; }
27 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
28 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
29 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
30 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;      }
31 template<class T> void SWAP(T& a, T& b) { T x = a; a = b; b = x; }
32 
33 //typedef __int64 LL;
34 typedef long long LL;
35 const int MAXN = 200010;
36 const int MAXM = 100005;
37 const double eps = 1e-10;
38 //const LL MOD = 1000000007;
39 
40 int step[1<<21];
41 bool vis[1<<21];
42 
43 int BFS(int s)
44 {
45     vis[s] = 1;
46     queue<int>q;
47     q.push(s);
48     step[s] = 0;
49     while(!q.empty())
50     {
51         int u = q.front(); q.pop();
52         if(u == 0)  return step[u];
53         for(int i=1;i<19;i++)
54         {
55             int r = u;
56             r ^= (1<<i-1) | (1<<i) | (1<<i+1);
57             if(!vis[r])
58             {
59                 vis[r] = 1;
60                 step[r] = step[u] + 1;
61                 if(r==0) return step[r];
62                 q.push(r);
63             }
64         }
65         int r = u ^ (1<<0) ^ (1<<1);
66         if(!vis[r]) { vis[r] = 1; step[r] = step[u] + 1; q.push(r); if(r==0) return step[r];}
67         r = u ^ (1<<18) ^ (1<<19);
68         if(!vis[r]) { vis[r] = 1; step[r] = step[u] + 1; q.push(r); if(r==0) return step[r];}
69     }
70     return -1;
71 }
72 
73 int main()
74 {
75         //FOPENIN("in.txt");
76         int st = 0, x;
77         for(int i=0;i<20;i++)
78         {
79             scanf("%d", &x);
80             st |= (x<<i);
81         }
82         printf("%d\n", BFS(st));
83         return 0;
84 }

View Code

然后突然想起之前寫過一個靜態隊列的模板，是用循環隊列寫的，想著正好去測試下，就改了隊列的定義，其他使用完全一致，沒任何修改，結果跑了250ms快了好多了啊有木有。。

  1 #include <map>
  2 #include <set>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <vector>
  8 #include <cstdio>
  9 #include <cctype>
 10 #include <cstring>
 11 #include <string.h>
 12 #include <cstdlib>
 13 #include <iostream>
 14 #include <algorithm>
 15 using namespace std;
 16 #define INF 1e9
 17 #define inf (-((LL)1<<40))
 18 #define lson k<<1, L, mid
 19 #define rson k<<1|1, mid+1, R
 20 #define mem0(a) memset(a,0,sizeof(a))
 21 #define mem1(a) memset(a,-1,sizeof(a))
 22 #define mem(a, b) memset(a, b, sizeof(a))
 23 #define FOPENIN(IN) freopen(IN, "r", stdin)
 24 #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
 25 template<class T> T CMP_MIN(T a, T b) { return a < b; }
 26 template<class T> T CMP_MAX(T a, T b) { return a > b; }
 27 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 28 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 29 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 30 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;      }
 31 template<class T> void SWAP(T& a, T& b) { T x = a; a = b; b = x; }
 32 
 33 //typedef __int64 LL;
 34 typedef long long LL;
 35 const int MAXN = 200010;
 36 const int MAXM = 100005;
 37 const double eps = 1e-10;
 38 //const LL MOD = 1000000007;
 39 
 40 
 41 //MyQueue<Type>q;
 42 //定義了一個固定長度的隊列， 不能動態增長
 43     //構造時不傳參數，隊列大小為1e5，傳入參數時為自定義大小
 44     //如果隊列不為空，front返回隊首元素，
 45     //如果隊列為空，pop無效，front返回NULL
 46     //clear將隊列清空, 供多次使用
 47     //如果push時產生沖突，即隊列已滿， 將加入失敗
 48 template <class T>
 49 class MyQueue
 50 {
 51 private:
 52     T* que;
 53     int si, fr, re;
 54     void setValue(int _size) {
 55         fr = 0; re = 0;
 56         si = _size;
 57         que = (T*)malloc(si * sizeof(T));
 58     }
 59 public:
 60     MyQueue() {
 61         this->setValue(100005);
 62     }
 63     MyQueue(int _size) {
 64         this->setValue(_size);
 65     }
 66     T front() {
 67         if(fr != re)
 68             return que[fr];
 69         return NULL;
 70     }
 71     void pop() {
 72         if(fr != re)
 73             fr = (fr + 1) % si;
 74     }
 75     void push(T e) {
 76         if((re + 1) % si == fr) return ;
 77         que[re] = e;
 78         re = (re + 1) % si;
 79     }
 80     bool empty() {
 81         if(fr == re) return 1;
 82         return 0;
 83     }
 84     void clear() {
 85         fr = 0;
 86         re = 0;
 87     }
 88 };
 89 
 90 int step[1<<21];
 91 bool vis[1<<21];
 92 
 93 int BFS(int s)
 94 {
 95     vis[s] = 1;
 96     MyQueue<int>q(1<<21);//定義隊列的大小為(1<<21),其他無任何修改
 97     q.push(s);
 98     step[s] = 0;
 99     while(!q.empty())
100     {
101         int u = q.front(); q.pop();
102         if(u == 0)  return step[u];
103         for(int i=1;i<19;i++)
104         {
105             int r = u;
106             r ^= (1<<i-1) | (1<<i) | (1<<i+1);
107             if(!vis[r])
108             {
109                 vis[r] = 1;
110                 step[r] = step[u] + 1;
111                 q.push(r);
112             }
113         }
114         int r = u ^ (1<<0) ^ (1<<1);
115         if(!vis[r]) { vis[r] = 1; step[r] = step[u] + 1; q.push(r); }
116         r = u ^ (1<<18) ^ (1<<19);
117         if(!vis[r]) { vis[r] = 1; step[r] = step[u] + 1; q.push(r); }
118     }
119     return -1;
120 }
121 
122 int main()
123 {
124         //FOPENIN("in.txt");
125         int st = 0, x;
126         for(int i=0;i<20;i++)
127         {
128             scanf("%d", &x);
129             st |= (x<<i);
130         }
131         printf("%d\n", BFS(st));
132         return 0;
133 }