BZOJ 1560 火星藏宝图(DP)

很容易想到直接排序然后DP。令dp[i]为到达i点的最大收益，则有dp[i]=max(dp[j]-(xi-xj)^2-(yi-yj)^2+v[i]).(j<=i,xj<=xi,yj<=yi)。

时间复杂度为O(n^2).显然超时。

考虑到转移的特性。假设在计算dp[i]的时候，j，k在同一列且xj<xk。那么一定有j通过k转移到i比直接从j转移更优。证明很容易。

所以我们只需要枚举小于等于i的列，取当前列的最大纵坐标转移即可。

时间复杂度O(nm).

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=200005;
//Code begin...

struct Node{int x, y, v;}node[N];
int vis[1005], dp[N];

bool comp(Node a, Node b){
    if (a.y==b.y) return a.x<b.x;
    return a.y<b.y;
}
int main ()
{
    int n, m;
    n=Scan(); m=Scan();
    FOR(i,1,n) node[i].x=Scan(), node[i].y=Scan(), node[i].v=Scan(), dp[i]=-INF;
    sort(node+1,node+n+1,comp);
    dp[1]=node[1].v; vis[1]=1;
    FOR(i,2,n) {
        for (int j=node[i].x; j>=1; --j) {
            if (!vis[j]) continue;
            int k=vis[j];
            dp[i]=max(dp[i],dp[k]-(node[i].x-node[k].x)*(node[i].x-node[k].x)-(node[i].y-node[k].y)*(node[i].y-node[k].y));
        }
        dp[i]+=node[i].v;
        vis[node[i].x]=i;
    }
    printf("%d\n",dp[n]);
    return 0;
}