版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/zjucor/article/details/84197963
Given an array A of strings, find any smallest string that contains each string in A as a substring.
We may assume that no string in A is substring of another string in A.
Example 1:
Input: ["alex","loves","leetcode"] Output: "alexlovesleetcode" Explanation: All permutations of "alex","loves","leetcode" would also be accepted.
Example 2:
Input: ["catg","ctaagt","gcta","ttca","atgcatc"] Output: "gctaagttcatgcatc"
Note:
1 <= A.length <= 121 <= A[i].length <= 20
思路：其实就是TSP问题，用状压DP。比赛的时候想到最小生成树上去了，尴尬
https://leetcode.com/articles/find-the-shortest-superstring/
#import heapq
class Solution:
def shortestSuperstring(self, A):
"""
:type A: List[str]
:rtype: str
"""
n=len(A)
adj=[[0 for _ in range(n)] for _ in range(n)]
def helper(s1,s2):
ma=min(len(s1),len(s2))
for k in range(ma,0,-1):
if s1[len(s1)-k:]==s2[:k]: return k
return 0
for i in range(n):
for j in range(n):
if i==j: continue
adj[i][j]=helper(A[i],A[j])
dp=[[-1 for _ in range(n)] for _ in range(2**n)]
path=[[-1 for _ in range(n)] for _ in range(2**n)]
for mask in range(2**n):
for bit in range(n):
if mask&(1<<bit):
mask2 = mask^(1<<bit)
if mask2==0:
dp[mask][bit]=0
continue
for i in range(n):
if mask2&(1<<i):
t = dp[mask2][i]+adj[i][bit]
if t>dp[mask][bit]:
dp[mask][bit] = t
path[mask][bit]=i
mask = 2**n-1
s = dp[mask].index(max(dp[mask]))
res=[]
while len(res)<n:
res.append(s)
s = path[mask][s]
mask = mask^(1<<res[-1])
res=res[::-1]
s = A[res[0]]
for i in range(1,n):
s+=A[res[i]][adj[res[i-1]][res[i]]:]
return s
s=Solution()
print(s.shortestSuperstring( ["yeeiebcz","qbqhdytk","ygueikth","thqzyeei","gyygueikt","ikthqzyee"]))
print(s.shortestSuperstring( ["alex","loves","leetcode"]))
#print(s.shortestSuperstring( ["catg","ctaagt","gcta","ttca","atgcatc"]))
#print(s.shortestSuperstring(["ift","efd","dnete","tef","fdn"]))