Codeforces 828 C String Reconstruction

题目地址：http://codeforces.com/contest/828/problem/C
题意：给你一些字符串的提示，让你拼接出一个完整的字典序最小的字符串（PS：是一定会有字符串的，所以重复的可以不用去看，因为一定是对的，我就是一开始没看懂这个TLE了）
思路：把起点的顺序排序，再遍历。重复的就不用考虑，详细看代码，比较容易理解，仔细点就好了

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <stack>#include <cmath>#include <cstdio>#include <algorithm>#define LL long long #define N 10000010#define M 50010#define inf 0x3f3f3f3fusing namespace std;const LL mod = 1e9 + 7;const double eps = 1e-9;string str[N];struct node {    int id, x;}now;vector<node>v;bool cmp(node a, node b) {    if (a.x == b.x) {        return str[a.id].length() > str[b.id].length();    }    return a.x < b.x;}int main() {    cin.sync_with_stdio(false);    int n, m;    while (cin >> n) {        v.clear();        for (int i = 0; i < n; i++) {            cin >> str[i] >> m;            now.id = i;            for (int j = 0; j < m; j++) {                cin >> now.x;                v.push_back(now);            }        }        sort(v.begin(), v.end(), cmp);        int len = v.size();        int num = 1;        for (int i = 0; i < len; i++) {            while (num < v[i].x) {                num++;                cout << "a";            }            if (num != v[i].x) {                if (num > v[i].x + str[v[i].id].length()) {                    continue;                }                for (int j = num - v[i].x; j < str[v[i].id].length(); j++) {                    cout << str[v[i].id][j];                    num++;                }            }            else {                cout << str[v[i].id];                num += str[v[i].id].length();            }        }        cout << endl;    }    return 0;}