Delicious Apples
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 321 Accepted Submission(s): 95
Problem Description
There are apple
trees planted along a cyclic road, which is metres
long. Your storehouse is built at position on
that cyclic road.
The th
tree is planted at position ,
clockwise from position .
There are delicious
apple(s) on the th
tree.
You only have a basket which can contain at most 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?
There are less than 20 huge testcases, and less than 500 small testcases.
The
You only have a basket which can contain at most
There are less than 20 huge testcases, and less than 500 small testcases.
Input
First line: ,
the number of testcases.
Then testcases
follow. In each testcase:
First line contains three integers, .
Next lines,
each line contains .
Then
First line contains three integers,
Next
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
解题思路:
注意到,最多仅仅有一次会绕整个圈走一次。因此,先贪心的处理左半环和右半环。然后枚举绕整圈的时候从左側摘得苹果和从右側摘得苹果的数目。
#include <iostream> #include <cstring> #include <cstdio> #include <cstdlib> #include <vector> #include <queue> #include <stack> #include <cmath> #include <algorithm> #define LL long long using namespace std; const int MAXN = 100000 + 10; int L, N, K; LL x[MAXN]; LL ld[MAXN], rd[MAXN]; vector<LL>l, r; int main() { int T; scanf("%d", &T); while(T--) { scanf("%d%d%d", &L, &N, &K); l.clear(); r.clear(); int pos, num, m = 0; for(int i=1;i<=N;i++) { scanf("%d%d", &pos, &num); for(int i=1;i<=num;i++) x[++m] = (LL)pos; } for(int i=1;i<=m;i++) { if(2 * x[i] < L) l.push_back(x[i]); else r.push_back(L - x[i]); } sort(l.begin(), l.end()); sort(r.begin(), r.end()); int lsz = l.size(), rsz = r.size(); memset(ld, 0, sizeof(ld)); memset(rd, 0, sizeof(rd)); for(int i=0;i<lsz;i++) ld[i + 1] = (i + 1 <= K ? l[i] : ld[i + 1 - K] + l[i]); for(int i=0;i<rsz;i++) rd[i + 1] = (i + 1 <= K ? r[i] : rd[i + 1 - K] + r[i]); LL ans = (ld[lsz] + rd[rsz]) * 2; for(int i=0;i<=lsz&&i<=K;i++) { int p1 = lsz - i; int p2 = max(0, rsz-(K-i)); ans = min(ans, 2*(ld[p1] + rd[p2]) + L); } cout << ans << endl; } return 0; }
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