Wooden Sticks
Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 11244 Accepted Submission (s): 4627
Problem Description There is a pile of n wooden sticks. The length and weight of each stick is known in advance. The sticks is processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times is associated with cleaning operations and changing tools and shapes. The setup times of the woodworking machine is given as follows:
(a) The setup time for the first wooden stick is 1 minute.
(b) Right after processing a stick of length L and weight W, the machine would need no setup time for a stick of length l ' and weight W ' if l<=l ' and W<=w '. Otherwise, it'll need 1 minute for setup.
You is to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight is (4,9), (5,2), (2,1), (3,5), and (1,4), then the Minimum setup time should be 2 minutes since there are a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).
Input the input consists of T test cases. The number of test cases (T) is given on the first line of the input file. Each test case consists of a lines:the first line has a integer n, 1<=n<=5000, that represents the number of Wo Oden sticks in the test case, and the second line contains n 2 positive integers L1, W1, L2, W2, ..., LN, WN, each of Magn Itude at most 10000, where Li and wi is the length and weight of the i th wooden stick, respectively. The 2n integers is delimited by one or more spaces.
Output the output should contain the minimum setup time in minutes, one per line.
Sample Input
3 5 4 9 5 2 2 1 3 5 1 4 3 2 2 1 1 2 2 3 1 3 2 2 3 1
Sample Output
2 1 3
Test instructions is the first 3 is 3 sets of data, then a 5 is 5 sticks, and then each input its length and weight. The question is how to do these sticks with the least time, when doing a stick, if the latter stick, length and weight are greater than the previous
One, you don't have to add time.
Idea..
Take the first set of data to say:
First sort by length in small to large: (1,4), (2,1), (3, 5), (4, 9), (5,2);
We found that (2, 1) The weight is too small, and (5, 2) The length is too long. The result of this row is 3;
In fact, we just need to mark up the unsatisfied first.
(1,4), (3,5), (4,9) ———— (2,1), (5,2)
But we have this line, which is the order of the topic, the result is 2;
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <
vector> #include <queue> #include <sstream> #include <cmath> using namespace std;
#define M 100500 struct Node {int lenth;
int weight;
int k;
} Q[m];
BOOL CMP (Node A, Node B) {if (a.lenth==b.lenth) return a.weight<b.weight;
else return a.lenth<b.lenth;
} int main () {int t;
int n;
scanf ("%d", &t);
while (t--) {scanf ("%d", &n);
for (int i=0; i<n; i++) {scanf ("%d%d", &q[i].lenth, &q[i].weight);
q[i].k= 1;
} sort (Q, q+n, CMP);
int ans = 0; for (int i=0; i<n-1; i++) {if (!
Q[I].K) continue;
int t = q[i].weight;
for (int j = i+1; j<n; j + +) if (q[j].weight>=t &&q[j].k==1) { ans++;
q[j].k=0;
t = q[j].weight;
}} printf ("%d\n", N-ans);
} return 0;
}