Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

A supermarket has a set Prod of products on sale. It earns a profit p_x for each product x in Prod sold by a deadline d_x that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell from Prod such that the selling of each product x in Sell, according to the ordering of Sell, completes before the deadline d_x or just when d_x expires. The profit of the selling schedule is . An optimal selling schedule is a schedule with a maximum profit.

For example, consider the products Prod={a,b,c,d} with (p_a,p_a=(50,2), (p_b,p_b=(10,1), (p_c,p_c=(20,2), and (p_d,p_d=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80.

Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products.

Input 

A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.

Output 

For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.

Sample Input 

4  50 2  10 1   20 2   30 1

7  20 1   2 1   10 3  100 2   8
2
5 20  50 10

Sample Output 

80
185

The sample input contains two product sets. The first set encodes the products from table 1. The second set is for 7 products. The profit of an optimal schedule for these products is 185.

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對價值作排序(由大到小)，使用 disjoint set 去完成模擬的動作，
盡可能放在靠近 deadline 的位置(也就是盡可能靠右)。
disjoint set 模擬兩區間(使用過)的合併，然後多紀錄一個最左邊的區間 l，
到時候查詢這個最左邊 l-1 放置這個選項，如果 l-1 <= 0 則棄置。

#include <stdio.h>
#include <algorithm>
using namespace std;
struct task {
    int d, p;
};
bool cmp(task a, task b) {
    return a.p > b.p;
}
int p[10005], r[10005], inter[10005];
void init(int n) {
    for(int i = 0; i <= n; i++)
        p[i] = i, r[i] = 1, inter[i] = i;
}
int findp(int x) {
    return p[x] == x ? x : (p[x]=findp(p[x]));
}
int joint(int x, int y) {
    x = findp(x), y = findp(y);
    if(x != y) {
        if(r[x] > r[y]) {
            p[y] = x, r[x] += r[y];
            inter[x] = min(inter[x], inter[y]);
        } else {
            p[x] = y, r[y] += r[x];
            inter[y] = min(inter[x], inter[y]);
        }
        return 1;
    }
    return 0;
}
int main() {
    int n, i, x, y;
    int sche[10005];
    while(scanf("%d", &n) == 1) {
        task A[10005];
        int day = 1, ans = 0;
        for(i = 0; i < n; i++) {
            scanf("%d %d", &A[i].p, &A[i].d);
            day = max(day, A[i].d);
        }
        day++;
        for(i = 1; i <= day; i++)
            sche[i] = 0;
        sort(A, A+n, cmp);
        init(day);
        for(i = 0; i < n; i++) {
            if(sche[A[i].d] == 0) {
                x = y = A[i].d;
                if(sche[x+1])
                    x++;
            } else {
                x = findp(A[i].d);
                y = inter[x]-1;
                if(!y)  continue;
            }
            sche[y] = i+1;
            ans += A[i].p;
            joint(x, y);
            if(y-1 > 0 && sche[y-1])
                joint(y-1, y);
        }
        printf("%d\n", ans);
    }
    return 0;
}
/*
4  50 2  10 1   20 2   30 1

7  20 1   2 1   10 3  100 2   8
2
5 20  50 10
*/