POJ 1795 DNA Laboratory (贪心+状压DP)

题意：给定 n 个 字符串，让你构造出一个最短，字典序最小的字符串，包括这 n 个字符串。

析：首先使用状压DP，是很容易看出来的，dp[s][i] 表示已经满足 s 集合的字符串以 第 i 个字符串结尾，他很容易就求得最短长度，但是这个字符串怎么构造呢，

由于要字典序最小，所以就不好搞了，挺麻烦的，所以我们利用贪心的思路，我们可以这样定义，dp[s][i] 表示已经满足 s 集合的字符串以 第 i 个字符串开头，

从后向前放，状态转移方程为：dp[s|(1<<i)][i] = min{ dp[s][j] + dist[k][j] }，dist[k][j] 表示把 k 放到 j 前面所要的最短长度，这个数组我们可以通过预处理来得到，

注意这里是把 k 放到 j 的前面，不是把 j 放到 k 的后面。最后还要注意把包含的字符串去掉，还有如果全为一样的情况，要注意特殊判断，这个点我RE了一晚上。。。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 100000000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[1<<16][16], dist[20][20];
string str[20], ans;
vector<string> v;

void init(){
  for(int i = 0; i < n; ++i)
    for(int j = 0; j < n; ++j) if(i != j){
      int t = min(v[i].size(), v[j].size());
      dist[i][j] = 0;
      for(int k = t; k >= 0; --k)
        if(v[i].substr(v[i].size() - k) == v[j].substr(0, k)){
          dist[i][j] = v[i].size() - k;
          break;
        }
    }
}

void dfs(int id, int s){
  if(s == 0)  return ;
  string ss = "Z";
  int x;
  for(int i = 0; i < n; ++i){
    if(!(s&(1<<i)))  continue;
    if(dp[s|(1<<id)][id] == dp[s][i] + dist[id][i]){
      int xx = v[id].size() - dist[id][i];
      string sss = v[i].substr(xx);
      if(ss > sss)  ss = sss, x = i;
    }
  }
  ans += ss;
  dfs(x, s^(1<<x));
}

int main(){
  ios_base::sync_with_stdio(false);
  int T;   cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    cin >> n;
    cout << "Scenario #" << kase << ":" << endl;
    for(int i = 0; i < n; ++i)  cin >> str[i];
    v.clear();
    for(int i = 0; i < n; ++i){
      bool ok = true;
      for(int j = 0; j < n; ++j){
        if(i == j || str[i].size() > str[j].size())  continue;
        if(str[j].find(str[i]) != string::npos){
          ok = false;
          break;
        }
      }
      if(ok)  v.push_back(str[i]);
    }
    if(v.empty()){
      cout << str[0] << endl << endl;
      continue;
    }
    sort(v.begin(), v.end());
    n = v.size();

    init();
    int all = 1<<n;
    memset(dp, INF, sizeof dp);
    for(int i = 0; i < n; ++i)
      dp[1<<i][i] = v[i].size();

    for(int i = 0; i < all; ++i)
      for(int j = 0; j < n; ++j){
        if(dp[i][j] == INF)  continue;
        for(int k = 0; k < n; ++k){
          if(i & (1<<k))  continue;
          dp[i|(1<<k)][k] = min(dp[i|(1<<k)][k], dp[i][j] + dist[k][j]);
        }
      }
    int id = 0;
    for(int i = 0; i < n; ++i)  if(dp[all-1][id] > dp[all-1][i])  id = i;
    ans = v[id];
    dfs(id, (all-1)^(1<<id));
    cout << ans << endl << endl;
  }
  return 0;
}