[离散化][前缀和][DP][二分]JZOJ 5873 小p的属性

Description

 

Input

Output

 

Sample Input

2 4
2 1 10
1 2 20
 

Sample Output

50

 

Data Constraint

分析

我们可以离散化之后处理二维前缀和（用二分找到位置）

然后DP方程显而易见

#include <iostream>
#include <cstdio>
#include <algorithm> 
using namespace std;
typedef long long ll;
const int N=1e3+10;
ll x[N],y[N],a[N][3];
ll s[N][N],f[N][N];
int n,m; 
int cnt,nx,ny;

int Bin_Search(int val,bool ot) {
    int l=1,r=ot?ny:nx;
    while (l<=r) {
        int mid=l+r>>1;
        if (ot?y[mid]==val:x[mid]==val) return mid;
        if (ot?y[mid]<val:x[mid]<val) l=mid+1;
        else r=mid-1;
    }
}

int main() {
    freopen("growth.in","r",stdin);
    freopen("growth.out","w",stdout);
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++) {
        int a1,a2,a3;
        scanf("%d%d%d",&a1,&a2,&a3);
        if (a1+a2>m) continue;
        a[++cnt][0]=a1;a[cnt][1]=a2;a[cnt][2]=a3;
        x[cnt]=a1;y[cnt]=a2;
    }
    sort(x+1,x+cnt+1);sort(y+1,y+cnt+1);
    nx=unique(x+1,x+cnt+1)-x-1;ny=unique(y+1,y+cnt+1)-y-1;
    for (int i=1;i<=cnt;i++)
        s[Bin_Search(a[i][0],0)][Bin_Search(a[i][1],1)]+=a[i][2];
    for (int i=1;i<=nx;i++)
        for (int j=1;j<=ny;j++)
            s[i][j]+=s[i-1][j]+s[i][j-1]-s[i-1][j-1];
    for (int i=1;i<=nx;i++)
        for (int j=1;j<=ny;j++)
            f[i][j]=s[i][j]+max(f[i-1][j]+(x[i]-x[i-1]-1)*s[i-1][j],f[i][j-1]+(y[j]-y[j-1]-1)*s[i][j-1]);
    ll lans=0;
    for (int i=1;i<=nx;i++)
        for (int j=1;j<=ny;j++)
            lans=max(lans,f[i][j]+(m-x[i]-y[j])*s[i][j]);
    printf("%lld",lans);
    fclose(stdin);fclose(stdout);
}

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