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】

 

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long LL;
const int maxn = 1000+10;
const int maxm = 200000+10;
const int INF = 1e9;

struct node{
    int x,y,p;
    bool operator<(const node& rhs) const{
        return (x<rhs.x)||(x==rhs.x && y<rhs.y);
    }
}a[maxm];

int pos[maxn],n,m;
LL d[maxn];

int read(int& x) {
    char c=getchar(); 
    while(!isdigit(c)) c=getchar();
    x=0;
    while(isdigit(c)) x=x*10+c-'0',c=getchar();
}

int main() {
    scanf("%d%d",&n,&m);
    n+=2;
    a[1].x=a[1].y=1;
    a[2].x=a[2].y=m;
    for(int i=3;i<=n;i++) 
        read(a[i].x),read(a[i].y),read(a[i].p);
    sort(a+2,a+n+1);
    pos[1]=1;
    for(int i=2;i<=n;i++) {
        LL tmp=-INF;
        for(int j=1;j<=a[i].y;j++) if(pos[j])
            tmp=max(tmp,d[j]-(a[i].y-j)*(a[i].y-j)-(a[i].x-pos[j])*(a[i].x-pos[j]));
        pos[a[i].y]=a[i].x , d[a[i].y]=tmp+a[i].p;
    }
    printf("%lld\n",d[m]);
    return 0;
}