HDU 5303 Delicious Apples (貪心 列舉 好題)

_TCgogogo_發表於2015-07-23
APP


Delicious Apples

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

Total Submission(s): 199    Accepted Submission(s): 54

Problem Description
There are n apple trees planted along a cyclic road, which is L metres long. Your storehouse is built at position 0 on that cyclic road.
The ith tree is planted at position xi, clockwise from position 0. There are ai delicious apple(s) on the ith tree.

You only have a basket which can contain at most K apple(s). You are to start from your storehouse, pick all the apples and carry them back to your storehouse using your basket. What is your minimum distance travelled?

1≤n,k≤105,ai≥1,a1+a2+...+an≤105
1≤L≤109
0≤x[i]≤L

There are less than 20 huge testcases, and less than 500 small testcases.
 
Input
First line: t, the number of testcases.
Then t testcases follow. In each testcase:
First line contains three integers, L,n,K.
Next n lines, each line contains xi,ai.
 
Output
Output total distance in a line for each testcase.
 
Sample Input
2
10 3 2
2 2
8 2
5 1
10 4 1
2 2
8 2
5 1
0 10000




 




Sample Output




18
26




 




Source



2015 Multi-University Training Contest 2


題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=5303


題目大意：有一個長為L的環，n個蘋果樹，一個籃子最多裝k個蘋果，裝完要回到起點卸下再出發，給出n個蘋果樹順時針的位置及蘋果的個數，求摘完所有蘋果走的最小路程


題目分析：顯然，只有在某種特殊條件下，即兩側都還有蘋果且可以一次裝完且最後的蘋果都離起點比較遠，這種情況下，我們直接繞圈可能會更優，也就是說整圈最多繞一次，因此我們可以先對兩邊貪心，題目的資料顯示蘋果的數量最多就1e5，顯然我們可以把蘋果“離散”出來，用x[i]記錄第i個蘋果到起點的位置，然後對位置從小到大排序，先選擇路程小的，選擇的時候用dis[i]記錄單側裝了i個蘋果的最小路程，類似揹包計數的原理，答案要乘2，因為是來回的，最後在k>=i時，列舉繞整圈的情況，szl－i表示只走左邊採的蘋果數，szr -
 (k - i)表示只走右邊採的蘋果樹，畫個圖就能看出來了，注意右邊這裡可能值為負，要和0取最大，然後答案就是（disl[szl－i] + disr[szr - (k - i)]）* 2 ＋L，這裡其實畫圖更加直觀。最後取最小即可，注意有幾個wa點，一個是要用long long，二是之前說的出現負數和0取大，三是每次要清零

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 1e5 + 5;
int  L, n, k;
ll x[MAX], disl[MAX], disr[MAX];
vector <ll> l, r;

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        memset(disl, 0, sizeof(disl));
        memset(disr, 0, sizeof(disr));
        l.clear();
        r.clear();
        scanf("%d %d %d", &L, &n, &k);
        int cnt = 1;
        for(int i = 1; i <= n; i++) 
        {
            ll pos, num;
            scanf("%lld %lld", &pos, &num);
            for(int j = 1; j <= num; j++)
                x[cnt ++] = (ll) pos;   //離散操作
        }
        cnt --;
        for(int i = 1; i <= cnt; i++)
        {
            if(2 * x[i] < L)
                l.push_back(x[i]);
            else
                r.push_back(L - x[i]);  //記錄位置
        }   
        sort(l.begin(), l.end());
        sort(r.begin(), r.end());
        int szl = l.size(), szr = r.size();
        for(int i = 0; i < szl; i++)
            disl[i + 1] = (i + 1 <= k ? l[i] : disl[i + 1 - k] + l[i]);
        for(int i = 0; i < szr; i++)
            disr[i + 1] = (i + 1 <= k ? r[i] : disr[i + 1 - k] + r[i]);
        ll ans = (disl[szl] + disr[szr]) * 2;
        for(int i = 0; i <= szl && i <= k; i++)
        {
            int p1 = szl - i;
            int p2 = max(0, szr - (k - i));
            ans = min(ans, 2 * (disl[p1] + disr[p2]) + L);
        }
        printf("%I64d\n", ans);
    }
}











            




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