题目大意
给你n个串和在原串中的出现位置，问原串
思路
直接跑肯定是GG
考虑怎么优化
因为保证有解，所以考虑过的点我们就不再考虑
用并查集维护当前每个点之后最早的没有被更新过的点
然后就做完了，很巧妙对吧
c++//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
//typename
typedef long long ll;
//convenient for
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
//inf of different typename
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
//fast read and write
template <typename T>
void Read(T &x) {
bool w = 1;x = 0;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-') w = 0, c = getchar();
while (isdigit(c)) {
x = (x<<1) + (x<<3) + c -'0';
c = getchar();
}
if (!w) x = -x;
}
template <typename T>
void Write(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) Write(x / 10);
putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = 1e6 + 10;
int n, fa[N << 1];
char s[N << 1], c[N];
void init() {
fu(i, 1, (N << 1) - 1) fa[i] = i;
}
int find(int x) {
return x == fa[x] ? x : fa[x] = find(fa[x]);
}
int main() {
Read(n);
init();
int maxl = 0;
fu(i, 1, n) {
scanf("%s", c + 1);
int num, len = strlen(c + 1), pos;
Read(num);
fu(j, 1, num) {
Read(pos);
maxl = max(maxl, pos + len - 1);
for (int k = find(pos); k <= pos + len - 1; k = find(k)){
s[k] = c[k - pos + 1];
fa[k] = k + 1;
}
}
}
fu(i, 1, maxl) if (!s[i]) s[i] = 'a';
printf("%s", s + 1);
return 0;
}