bzoj 1560 [JSOI2009]火星藏宝图（DP）

1560: [JSOI2009]火星藏宝图

Time Limit: 10 Sec Memory Limit: 64 MB
Submit: 647 Solved: 309
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Description

Input

Output

Sample Input

4 10

1 1 20

10 10 10

3 5 60

5 3 30

Sample Output

-4

HINT

Source

JSOI2009Day2

【代码】

1 #include<cstdio>
2 #include<iostream>
3 #include<algorithm>
4 using namespace std;
5
6 typedef long long LL;
7 const int maxn = 1000+10;
8 const int maxm = 200000+10;
9 const int INF = 1e9;
10
11 struct node{
12     int x,y,p;
13     bool operator<(const node& rhs) const{
14         return (x<rhs.x)||(x==rhs.x && y<rhs.y);
15     }
16 }a[maxm];
17
18 int pos[maxn],n,m;
19 LL d[maxn];
20
21 int read(int& x) {
22     char c=getchar();
23     while(!isdigit(c)) c=getchar();
24     x=0;
25     while(isdigit(c)) x=x*10+c-'0',c=getchar();
26 }
27
28 int main() {
29     scanf("%d%d",&n,&m);
30     n+=2;
31     a[1].x=a[1].y=1;
32     a[2].x=a[2].y=m;
33     for(int i=3;i<=n;i++)
34         read(a[i].x),read(a[i].y),read(a[i].p);
35     sort(a+2,a+n+1);
36     pos[1]=1;
37     for(int i=2;i<=n;i++) {
38         LL tmp=-INF;
39         for(int j=1;j<=a[i].y;j++) if(pos[j])
40             tmp=max(tmp,d[j]-(a[i].y-j)*(a[i].y-j)-(a[i].x-pos[j])*(a[i].x-pos[j]));
41         pos[a[i].y]=a[i].x , d[a[i].y]=tmp+a[i].p;
42     }
43     printf("%lld\n",d[m]);
44     return 0;
45 }