B |
Mega Man’s Missions | |
|
Input |
Standard Input | |
|
Output |
Standard Output | |
Mega Man is off to save the world again. His objective is to kill the Robots
created by Dr. Wily whose motive is to conquer the world. In each mission, he
will try to destroy a particular Robot. Initially, Mega Man is equipped with a
weapon, called the “Mega Buster” which can be used to destroy the
Robots. Unfortunately, it may happen that his weapon is not capable
of taking down every Robot. However, to his fortune, he is capable of using the
weapons from Robots which he has completely destroyed and
these weapons maybe able to take down Robots which he otherwise cannot with his
own weapon. Note that, each of these enemy Robots carry exactly one weapon
themselves for fighting Mega Man. He is able to take down the Robots
in any order as long as he has at least one weapon capable of destroying the
Robot at a particular mission. In this problem, given the information about the
Robots and their weapons, you will have to determine the number of ways Mega Man
can complete his objective of destroying all the Robots.
Input
Input starts with an
integer T(T≤50),
the number of test cases.
Each test case starts with an
integer N(1≤N≤16).
Here N denotes the number of Robots to be
destroyed (each Robot is numbered from 1 to N).
This line is followed by N+1 lines, each
containing N characters. Each character will
either
be ‘1’ or ‘0’.
These lines represent
a (N+1)*N matrix. The rows are
numbered from 0 to N while the columns are
numbered from 1 to N. Row 0 represents the
information about the “Mega Buster”.
The jth character of
Row 0 will
be ‘1’ if
the “Mega Buster” can destroy
the jthRobot. For the
remaining N rows,
the jth character
of ith row will
be ‘1’ if the
weapon of ith Robot can destroy
the jth Robot. Note that, a
Robot’s weapon could be used to destroy the Robot itself, but this will have no
impact as the Robot must be destroyed anyway for its weapon to be acquired.
Output
For each case of input, there will be one line of output. It will first
contain the case number followed by the number of ways Mega Man can complete his
objective. Look at the sample output for exact format.
Sample Input |
Sample Output |
|
3 1 1 1 2 11 01 10 3 110 011 100 000 |
Case 1: 1 Case 2: 2 Case 3: 3 |
设dp[s] 为 已经摧毁掉的机器人的集合s的方法数
swep[s] 为摧毁机器人的集合s所拥有的武器集合
则dp[s] += dp[s ^ ( 1 << j]] ( 1 << j & j) && (
(1 << j) & （swep[ s ^ ( 1 << j)]) ( 0 =< j <
N)
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <bitset>
6
7 using namespace std;
8
9 typedef long long ll;
10 const int MAX_N = 20;
11 int N;
12 int wea[MAX_N],swea[1 << 17];
13 ll dp[1 << 17];
14
15
16 void init() {
17 for(int i = 0; i < (1 << N); ++i) {
18 swea[i] = wea[0];
19 for(int j = 0; j < N; ++j) {
20 if(1 << j & i) {
21 swea[i] |= wea[j + 1];
22 }
23 }
24 }
25
26 }
27 void solve() {
28 dp[0] = 1;
29 for(int i = 1; i <= N; ++i) {
30 int comb = (1 << i) - 1;
31 while(comb < 1 << N) {
32
33 for(int j = 0; j < N; ++j) {
34 if((1 << j & comb) && ((1 << j)
35 & (swea[comb ^ ( 1 << j)]))) {
36 dp[comb] += dp[comb ^ (1 << j)];
37 }
38 }
39 int x = comb & -comb, y = comb + x;
40 comb = ((comb & ~y) / x >> 1) | y;
41 }
42 }
43
44 printf("%lld\n",dp[(1 << N) - 1]);
45 }
46
47 int main()
48 {
49 // freopen("sw.in","r",stdin);
50 int t;
51 scanf("%d",&t);
52 int ca = 1;
53 while(t--) {
54 memset(dp,0,sizeof(dp));
55 memset(wea,0,sizeof(wea));
56 scanf("%d",&N);
57 for(int i = 0; i <= N; ++i) {
58 char s[20];
59 scanf("%s",s);
60 for(int j = 0; s[j] != ‘\0‘; ++j) {
61 if(s[j] != ‘0‘) wea[i] |= (1 << j);
62 }
63 }
64
65 init();
66 printf("Case %d: ",ca++);
67 solve();
68 }
69 //cout << "Hello world!" << endl;
70 return 0;
71 }
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