UVa 11100 - The Trip, 2007 难度: 0

题目

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2674

题意

n个正整数，尽量均匀地把它们分成严格递增序列。要求序列个数最小

思路

明显，序列个数就是出现最多次的正整数的次数。接下来只要循环着由小到大把数字放进各个序列中就可以了。

感想：

读了好几遍都没读懂题目“While maintaining the minimal number of pieces you are also to minimize the total number of bags in any one piece that must be carried”

代码

#include <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <tuple>
#define LOCAL_DEBUG
using namespace std;
typedef pair<int, int> MyPair;
const int MAXN = 1e4 + ;
const int MAXA = 1e6 + ;
MyPair a[MAXN];
int cnt[MAXA];
int cap[MAXA];

int main() {
#ifdef LOCAL_DEBUG
    freopen("C:\\Users\\Iris\\source\\repos\\ACM\\ACM\\input.txt", "r", stdin);
    //freopen("C:\\Users\\Iris\\source\\repos\\ACM\\ACM\\output.txt", "w", stdout);
#endif // LOCAL_DEBUG
    int T, n;
    for (int ti = ;cin>>n && n; ti++) {
        memset(cnt, , sizeof cnt);
        for (int i = ; i < n; i++) {
            int tmp;
            cin >> tmp;
            cnt[tmp] ++;
        }
        int ans = ;
        int newn = ;
        for (int i = ; i < MAXA; i++) {
            ans = max(ans, cnt[i]);
        }
        int cap_all = (n + ans - ) / ans;
        for (int i = ; i <= ans; i++) {
            cap[i] = cap_all;
        }
        int cid = ;
        int cidnow = ;
        for (int i = ; i < MAXA; i++) {
            ans = max(ans, cnt[i]);
            while (cap[cid] == ) {
                cid++;
            }
            while (cnt[i]) {
                a[newn].first = cidnow;
                a[newn].second = i;
                newn++;
                cnt[i]--;
                cap[cidnow]--;
                cidnow=(cidnow - cid + ) % (ans - cid)  + cid;
            }
        }
        sort(a, a + n);
        if (ti > )cout << endl;
        cout << ans << endl;
        for (int i = ; i < n; i++) {
            if (i && a[i].first != a[i - ].first)cout << endl;
            else if (i)cout << " ";
            cout << a[i].second;
        }
        cout << endl;
    }

    return ;
}