Leetcode 943. Find the Shortest Superstring(DP)


题目来源:https://leetcode.com/problems/find-the-shortest-superstring/description/

标记难度:Hard

提交次数:3/4

代码效率:2.93% -> 79.31%

题意

有N个字符串,找到最小的字符串S,使得这N个字符串都是S的子串。其中N<=12,字符串的长度<=20。

分析

这道题比赛的时候我没有做出来,但我自认为自己已经找到了正确的解法(确实差不多是正确的),只要再调一小会就能调出来了!结果事实是又花了两天才弄出来。我犯了这些错误:

  • 在搞错了状态变量的范围的同时没有设置好变量的初值
  • 计算两个字符串的overlap的函数少考虑了一种情况

所以就这样了……


我觉得比较简单的方法还是状态压缩DP。[1]dp[mask][i]表示总共包含mask这些字符串,且以A[i]作为结尾的字符串的最小长度(或者最大overlap长度;当字符串都是那么多时,这两者是一样的。然后就可以递推了:dp[mask ^ 1<<j][j] = max(dp[mask][i] + overlap(i, j))。显然,我们事实上可以不用保存具体的字符串(因为有最后一个字符串就够用了),而且可以事先计算出每两个字符串之间的overlap(这样就不需要重复计算)。不过这样就需要最后重建DP过程了……不过字符串处理过程太耗时了,也可以理解……

不过这样做了之后时间效率大大提高了(从1324ms提高到了28ms)

代码

特别慢的那个就不贴了……

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class Solution {
private:
inline int calcOverlap(const string& s1, const string& s2) {
int start = s1.length() - s2.length();
start = max(0, start);
for (int i = start; i < s1.length(); i++) {
int len = s1.length() - i;
if (s1.substr(i, len) == s2.substr(0, len))
return len;
}
return 0;
}

inline bool contains(int cnt, int i) {
return (cnt & (1 << i)) != 0;
}

public:
string shortestSuperstring(vector<string>& A) {
int n = A.size();
if (n == 1) return A[0];

// pre-calculate overlap graph
int overlap[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
overlap[i][j] = calcOverlap(A[i], A[j]);

const int MAX_CNT = (1 << n);
int f[MAX_CNT][n], parent[MAX_CNT][n];
for (int i = 0; i < n; i++) {
f[1 << i][i] = 0;
parent[1 << i][i] = -1;
}

// start DP
int ans = -1;
int p = -1;
for (int cnt = 2; cnt <= n; cnt++) {
for (int i = 0; i < MAX_CNT; i++) {
if (__builtin_popcount(i) != cnt) continue;
// ends with j
for (int j = 0; j < n; j++) {
if (!contains(i, j)) continue;
f[i][j] = -1;

int nmask = i ^ (1 << j);
// last one ends with k
for (int k = 0; k < n; k++) {
if (!contains(nmask, k)) continue;
if (f[nmask][k] + overlap[k][j] > f[i][j]) {
f[i][j] = f[nmask][k] + overlap[k][j];
parent[i][j] = k;
}
}

if (cnt == n && f[i][j] > ans) {
ans = f[i][j];
p = j;
}
}
}
}

// rebuild the path
string str;
int nmask = MAX_CNT - 1;
while (true) {
int par = parent[nmask][p];
if (par == -1) {
str = A[p] + str;
break;
}
str = A[p].substr(overlap[par][p], A[p].length() - overlap[par][p]) + str;
nmask ^= (1 << p);
p = par;
}

return str;
}
};

  1. Leetcode Official Solution for 943. Find the Shortest Superstring


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