Resource limit
Time limit: 1.0s Memory limit: 256.0MB
Problem description
Given the coordinates l, r and their value v of the left and right endpoints of n line segments on the number line, please select a number of line segments without a common point (the endpoints Coincidence is also considered to have a common point), so that their value sum is the largest, and the maximum value sum is output.
Input format
The first line is a positive integer n.
Next n lines, each line of three integers l, r, v represents the left endpoint, right endpoint and value of a line segment, respectively. l<r,v>0.
Output Format
Output an integer representing the maximum value and sum.
Sample input
4
1 3 4
3 5 7
5 7 3
2 6 8
Sample output
8
Data size and convention
n<=2000
l,r,v<=1000000
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Refer to the original text
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
public class Main {
static int[] dp = new int[1000005];
static Edge[] edges = new Edge[2005];
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for (int i = 0; i < n; i++) {
int l = sc.nextInt();
int r = sc.nextInt();
int v = sc.nextInt();
edges[i] = new Edge(l, r, v);
}
Arrays.sort(edges, 0, n, new Comparator<Edge>() {
@Override
public int compare(Edge o1, Edge o2) {
if (o1.r != o2.r) {
return o1.r - o2.r;
} else {
return o1.l - o2.l;
}
}
});
dp[edges[0].r] = edges[0].v;
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (edges[i].l <= edges[j].r){
dp[edges[i].r] = Math.max(dp[edges[i].r], Math.max(edges[i].v, dp[edges[j].r]));
} else {
dp[edges[i].r] = Math.max(dp[edges[i].r], edges[i].v + dp[edges[j].r]);
}
}
}
System.out.println(dp[edges[n-1].r]);
}
}
class Edge {
int l;
int r;
int v;
public Edge(int l, int r, int v) {
super();
this.l = l;
this.r = r;
this.v = v;
}
}
