UVa 11795 - Mega Man's Mission（集合DP 状态压缩）

Mega Man's Mission

Time Limit: 1000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

 

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 hascompletely 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). HereN denotes the number of Robots to be destroyed (each Robot is numbered from 1 toN). This line is followed by N+1 lines, each containingN characters. Each character will either be ‘1’ or‘0’. These lines represent a (N+1)*N matrix. The rows are numbered from 0 toN while the columns are numbered from 1 to N. Row 0 represents the information about the “Mega Buster”. Thej^th character of Row 0 will be‘1’if the “Mega Buster” can destroy the j^th Robot. For the remainingN rows, the j^th character ofi^th row will be ‘1’ if the weapon ofi^th Robot can destroy the j^th 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

 

 

题意：

最初只有一个武器，ｎ个机器人，每个机器人有一个武器，每个武器能消灭特定的一些机器人，消灭一个机器人就能得到他的武器。问消灭所有机器人的顺序总数。

dp[s] 表示消灭ｓ这个集合中机器人的方法数。

dp[s] 这个状态可以由其它状态通过消灭机器人ｉ得到。也就是

dp[s] = sum { dp[s^(1<<i)] }; ｉ为所有s^(1<<i)（表示还没有杀死ｉ）这个集合能够杀死并且在ｓ这个集合里面所有机器人; 

dp[s] 中的ｓ表示的是几经杀死ｓ这个集合中的机器人方法数

canKill[s] 表示杀死s这个集合中的机器人后能够杀死的机器人集合

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define P64I1(a) printf(oform1, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxn = 16 + 4;const int maxs = 1<<17;int A[maxn];int canKill[maxs];LL dp[maxs];int n;LL ans;char s[20];int main() {    int T;    scanf("%d", &T);    for(int kase=1; kase<=T; kase++) {        scanf("%d", &n);        int S0 = 0;        scanf("%s", s);        for(int i=0; i<n; i++) {            if(s[i]-'0') S0 |= 1<<i;        }        for(int i=0; i<n; i++) {            A[i] = 0;            scanf("%s", s);            for(int j=0; j<n; j++) {                if(s[j]-'0') A[i] |= 1<<j;            }        }        int maxState = (1<<n) - 1;        for(int s=0; s<=maxState; s++) {            canKill[s] = S0;            for(int i=0; i<n; i++) if(s&(1<<i))                canKill[s] |= A[i];        }        dp[0] = 1;        for(int s=1; s<=maxState; s++) {            dp[s] = 0;            for(int i=0; i<n; i++) if((s&(1<<i)) && (canKill[s^(1<<i)]&(1<<i))) {                dp[s] += dp[s^(1<<i)];            }        }        printf("Case %d: ", kase);        P64I(dp[maxState]);    }    return 0;}