[Codeforces Round #101 (Div. 2) C. Queue] STL之vector

[Codeforces Round #101 (Div. 2) C. Queue] STL之vector

题目链接：[Codeforces Round #101 (Div. 2) C. Queue]
题意描述：有N个人排队，现在告诉你每个人前面有ai个人身高比他高。(1≤N≤3000,0≤ai≤N−1)将队伍打乱之后，现在要你还原这个队伍。并且输出一种可能的结果，即每个人的身高hi(1≤hi≤109)。
解题思路：
引用官方题解：

Let’s sort the pairs (namei, ai) by ascending of ai. If there is an index i: 0 ≤ i < n that ai > i, then answer is “-1”. Otherwise the answer exists. We will iterate through the array of sorted pairs from left to right with supporting of vector of results res. Let on the current iteration ai = n - i, then we must transfer the current man in the position ai. It can be done in C++ with one line: res.insert(res.begin() + a[i], man);

#include <bits/stdc++.h>using namespace std;#define FIN             freopen("input.txt","r",stdin)#define FOUT            freopen("output.txt","w",stdout)#define fst             first#define snd             secondtypedef __int64 LL;//typedef long long LL;typedef pair<int, int> PII;const int MAXN = 3000 + 5;int N, M, K;char buf[MAXN][15];PII ele[MAXN];vector<PII> res;int main() {#ifndef ONLINE_JUDGE    FIN;#endif // ONLINE_JUDGE    while (~scanf ("%d", &N) ) {        for (int i = 1; i <= N; i++) {            scanf ("%s %d", buf[i], &ele[i].fst);            ele[i].snd = i;        }        sort (ele + 1, ele + N + 1);        bool suc = true;        for (int i = 1; i <= N; i++) {            if (ele[i].fst >= i) suc = false;        }        if (!suc) {            printf ("-1\n");            continue;        }        res.clear();        for (int i = 1, k = N; i <= N; i++, k--) {            res.insert (res.begin() + ele[i].fst, make_pair (ele[i].snd, k) );        }        for (int i = 0; i < N; i++) {            PII& e = res[i];            printf ("%s %d\n",  buf[e.fst], e.snd);        }    }    return 0;}