ZOJ 2563 Long Dominoes(状压DP)

给定一个m*n的方格子,要求用3*1的骨牌去覆盖,骨牌可以用横放或者竖放,问最终有多少种放置方式,将其铺满。

分析:由于最多30行,每行最多9列,所以可以按行来dp,设计每行的状态从而进行转移,考虑每个骨牌放置对下一行的影响,共有0,1,2,3种方式,0对应横放或者竖放时最下面那

个格子,此行对下一行没有影响,1,竖放时第1个,2竖放时第2个,这样进行转移。注意,第i行横放时要求上一行相应位置状态为0。

思路及代码都来自这里,其实不会做这题,看了才了解。

代码:

 1 #include <bits/stdc++.h>
 2 #define pb push_back
 3 #define mp make_pair
 4 #define esp 1e-14
 5 #define lson   l, m, rt<<1
 6 #define rson   m+1, r, rt<<1|1
 7 #define sz(x) ((int)((x).size()))
 8 #define pf(x) ((x)*(x))
 9 #define pb push_back
10 #define in freopen("solve_in.txt", "r", stdin);
11 #define bug(x) printf("Line : %u >>>>>>\n", (x));
12 #define inf 0x0f0f0f0f
13 using namespace std;
14 typedef long long LL;
15 typedef map<LL, int> MPS;
16 typedef pair<LL, LL> PII;
17 const int maxn = 20000;
18 LL dp[33][maxn];
19 int st[10];
20 int n, m;
21 void pre() {
22     st[0] = 1;
23     for(int i = 1; i <= 9; i++)
24         st[i] = st[i-1]*3;
25 }
26 int dig[20];
27 void getDig(int x) {
28     int len = 0;
29     memset(dig, 0, sizeof dig);
30     while(x) {
31         dig[len++] = x%3;
32         x /= 3;
33     }
34 }
35 void dfs(int cur, int state, int base, int prestate) {
36     if(cur == n) {
37         dp[base][state] += dp[base-1][prestate];
38         return;
39     }
40     if(dig[cur] == 0) {
41         if(cur+2 < n && dig[cur+1] == 0 && dig[cur+2] == 0) {
42             dfs(cur+3, state, base, prestate);
43         }
44         dfs(cur+1, state+st[cur], base, prestate);
45     } else if(dig[cur] == 1) {
46         dfs(cur+1, state+2*st[cur], base, prestate);
47     } else {
48         dfs(cur+1, state, base, prestate);
49     }
50 }
51 int main() {
52 //    in
53     pre();
54     while(scanf("%d%d", &n, &m), n || m) {
55         if(n*m%3) {
56             puts("0");
57             continue;
58         }
59         memset(dp, 0, sizeof dp);
60         dp[0][0] = 1;
61         for(int i = 1; i <= m; i++) {
62             for(int j = 0; j < st[n]; j++) {
63                 if(dp[i-1][j]) {
64                     getDig(j);
65                     dfs(0, 0, i, j);
66                 }
67             }
68         }
69         printf("%lld\n", dp[m][0]);
70     }
71     return 0;
72 }
View Code

 另一道类似的题目:HDU 4804 Campus Design

题意是:n*m的格子,用1*1和1*2的骨牌覆盖,但是有一些格子不需要覆盖,而且最终要求所用的1*1的骨牌个数num满足:c <= num <= d.

解法类似,每个格子被覆盖的情况有两种,0和1,即1*2格子竖放的第一个,1*2格子竖放第2个或者1*1格子,或者不需要覆盖,这样按行进行转移就行了。

代码:

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <map>
 7 #include <vector>
 8 #define pb push_back
 9 #define mp make_pair
10 #define esp 1e-14
11 #define lson   l, m, rt<<1
12 #define rson   m+1, r, rt<<1|1
13 #define sz(x) ((int)((x).size()))
14 #define pf(x) ((x)*(x))
15 #define pb push_back
16 #define in freopen("solve_in.txt", "r", stdin);
17 #define bug(x) printf("Line : %u >>>>>>\n", (x));
18 #define inf 0x0f0f0f0f
19 using namespace std;
20 typedef long long LL;
21 typedef map<LL, int> MPS;
22 typedef pair<LL, LL> PII;
23 
24 const int M = (int)1e9 + 7;
25 const int maxn = 110;
26 const int maxm = 1111;
27 
28 LL dp[maxn][23][maxm];
29 int n, m, c, d;
30 char maze[maxn][11];
31 int dig[11];
32 
33 void getDig(int x) {
34     memset(dig, 0, sizeof dig);
35     int len = 0;
36     while(x) {
37         dig[len++] = x&1;
38         x >>= 1;
39     }
40 }
41 bool check(char *s, int x) {
42     for(int i = 0; s[i]; i++) {
43         if((x&1) && s[i] == '0') return false;
44         x >>= 1;
45     }
46     return true;
47 }
48 void dfs(int pre, int num, int st, int cur, int state, int cnt) {
49     if(cnt > d) return;
50     if(cur == m) {
51         dp[pre+1][cnt][state] = (dp[pre+1][cnt][state]+dp[pre][num][st])%M;
52         return;
53     }
54     if(maze[pre+1][cur] == '0') {
55         dfs(pre, num, st, cur+1, state, cnt);
56     } else if(dig[cur] == 0) {
57         if(cur+1 < m && dig[cur+1] == 0 && maze[pre+1][cur+1] == '1')
58             dfs(pre, num, st, cur+2, state, cnt);
59         if(pre+1 < n && maze[pre+2][cur] == '1')
60         dfs(pre, num, st, cur+1, state+(1<<cur), cnt);
61         dfs(pre, num, st, cur+1, state, cnt+1);
62     } else {
63         dfs(pre, num, st, cur+1, state, cnt);
64     }
65 }
66 int main() {
67     
68     while(scanf("%d%d%d%d", &n, &m, &c, &d) == 4) {
69         memset(dp, 0, sizeof dp);
70         for(int i = 1; i <= n; i++)
71             scanf("%s", maze[i]);
72         dp[0][0][0] = 1;
73         for(int i = 0; i < n; i++) {
74             int j;
75             if(i == 0) {
76                 j = 0;
77                 getDig(j);
78                 int x = 0;
79                 if(dp[i][x][j])
80                     dfs(i, x, j, 0, 0, x);
81             } else {
82                 for(int j = 0; j < (1<<m);  j++) {
83                     if(check(maze[i], j) == false) continue;
84                     getDig(j);
85                     for(int x = 0; x <= d; x++) if(dp[i][x][j])
86                             dfs(i, x, j, 0, 0, x);
87                 }
88             }
89         }
90         LL ans = 0;
91         for(int i = c; i <= d; i++)
92             ans = (ans + dp[n][i][0])%M;
93         printf("%lld\n", ans);
94     }
95     return 0;
96 }
View Code

 

你可能感兴趣的:(long)