ural 2021 Scarily interesting!（贪心）

2021. Scarily interesting!

Time limit: 1.0 second
Memory limit: 64 MB

This year at Monsters University it is decided to arrange Scare Games. At the Games all campus gathers at the stadium stands, and the Scare program students divide into two teams to compete in their abilities of scaring children. This year the two teams will be “Oozma Kappa” and “Roar Omega Roar”.

Each team has n monsters, and the Games consist of n challenges. During each challenge Dean Hardscrabble, the chair of the Scare program, invites one monster from each team to demonstrate his mastery. Each of the monsters is invited only once and scores from 0 to 6 points, depending on how much a child is scared. The results of each challenge are announced at the same time for both monsters right after the end of this challenge. The winning team will be identified by the sum of the points scored by all its members.

Sports competition is an unpredictable process. But the Dean wants to keep all the course of the Games under control, so that the identity of the winning team will have been unclear for the audience as long as possible. For example, if six challenges until the end “Oozma Kappa” is forty points ahead, the audience at the stadium stands will just lose interest to the game. The Dean knows the skill level of all her students, and she wants to decide beforehand the order in which both teams’ members will be participating in the challenges. In what order should monsters from “Oozma Kappa” and from “Roar Omega Roar” show up to keep the audience in suspense as long as possible?

Input

The first line contains an integer n (2 ≤ n ≤ 1 000). The second line contains n integers within the range from 0 to 6, which are the points monsters from “Oozma Kappa” will score. The third line contains the points, monsters from “Roar Omega Roar” will score, written in the same manner.

Output

Output n lines, each containing integers oi and ri, which are the numbers of monsters from “Oozma Kappa” and “Roar Omega Roar” respectively, who should be called by the Dean to take part in the i-th challenge. In each team monsters are numbered with integers from 1 to n in the order they appear in the input data. If the problem has several solutions, output any of them.

Sample

input	output
5 0 1 4 3 6 6 5 1 3 0	5 1 1 5 4 4 2 3 3 2

Problem Author: Oleg Dolgorukov
Problem Source: NEERC 2014, Eastern subregional contest

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Difficulty: 298    Printable version    Submit solution    Discussion (3)
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两个队伍比赛，每个队伍有n个人，每个人的得分为0-6，总得分多的队伍赢。

求两个队伍怎样安排上场顺序，尽量让悬念保持到最后。

思路：赢的一方从小到大输出，输的一方从大到小输出。

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1024;

vector<int> vt1[7];
vector<int> vt2[7];

vector<int> vt3;
vector<int> vt4;

int main()
{
    int n;
    int i;
    int a, b;
    vector<int>::iterator it1, it2;
    int sum1, sum2;

    while (~scanf("%d", &n)) {
        for (i = 0; i <= 6; ++i) {
            vt1[i].clear();
            vt2[i].clear();
        }
        vt3.clear();
        vt4.clear();
        sum1 = 0;
        sum2 = 0;
        for (i = 1; i <= n; ++i) {
            scanf("%d", &a);
            vt1[a].push_back(i);
            sum1 += a;
        }
        for (i = 1; i <= n; ++i) {
            scanf("%d", &b);
            vt2[b].push_back(i);
            sum2 += b;
        }
//        printf("debug 1
");
        for (i = 0; i <= 6; ++i) {
            it1 = vt1[i].begin();
            it2 = vt2[i].begin();
            if (vt1[i].size() < vt2[i].size()) {
                while (it1 < vt1[i].end()) {
                    printf("%d %d
", *it1++, *it2++);
                }
                while (it2 < vt2[i].end()) {
                    vt4.push_back(*it2++);
                }
            } else {
//                printf("debug 3
");
                while (it2 < vt2[i].end()) {
                    printf("%d %d
", *it1++, *it2++);
                }
                while (it1 < vt1[i].end()) {
                    vt3.push_back(*it1++);
                }
            }
        }
//        printf("debug 2
");

        if (sum1 > sum2) {
            it1 = vt3.begin();
            it2 = vt4.end() - 1;
            while (it1 < vt3.end()) {
                printf("%d %d
", *it1++, *it2--);
            }
        } else {
            it1 = vt3.end() - 1;
            it2 = vt4.begin();
            while (it2 < vt4.end()) {
                printf("%d %d
", *it1--, *it2++);
            }
        }

    }

    return 0;
}

ps：自己没想到这样做啊，

我的思路是让双方的分差越小越好，但是不知道这样做对不，写起来也比较复杂。