bzoj 3156 防御准备（斜率优化+DP）

来源：互联网发布：java用redis缓存的步骤编辑：程序博客网时间：2024/06/05 05:13

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思路：

f[i] 表示前i个的最小花费　

转移：f[i] = f[j] + (i-j)*(i-j-1)/2 + A[i];

需要注意的是过程中数据超范围

代码一：

#include <bits/stdc++.h>using namespace std;typedef long long ll;#define mem(a) memset(a,0,sizeof(a))#define mp(x,y) make_pair(x,y)const int INF = 0x3f3f3f3f;const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}//////////////////////////////////////////////////////////////////////////const int maxn = 1e6+10;int n;ll A[maxn],f[maxn],q[maxn];double slope(ll j,ll k){return (1.0*(f[j]-f[k]-k*(k+1)/2+j*(j+1)/2))/(j-k);}int main(){n = read();for(int i=1; i<=n; i++)A[i] = read();int L=0,R=0;for(int i=1; i<=n; i++){while(L<R && slope(q[L+1],q[L])<i) L++;ll j = q[L];f[i] = f[j] + (i-j)*(i-j-1)/2 + A[i];while(L<R && slope(i,q[R])<slope(q[R],q[R-1])) R--;q[++R] = i;}cout << f[n] << endl;    return 0;}

代码二：

#include <bits/stdc++.h>using namespace std;typedef long long ll;#define mem(a) memset(a,0,sizeof(a))#define mp(x,y) make_pair(x,y)const int INF = 0x3f3f3f3f;const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}//////////////////////////////////////////////////////////////////////////const int maxn = 1e6+10;struct node{ll x,y;}now,q[maxn];int n;ll A[maxn];ll cross(node a,node b,node c){return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);}int main(){n = read();for(int i=1; i<=n; i++)A[i] = read();int L=0,R=0;for(int i=1; i<=n; i++){while(L<R && q[L+1].y-i*q[L+1].x <= q[L].y-i*q[L].x) L++;now.x = i;now.y = q[L].y-(ll)i*q[L].x+(ll)i*i+A[i];while(L<R && cross(q[R-1],q[R],now)<=0) R--;q[++R] = now;}cout << q[R].y-(ll)n*(n+1)/2 << endl;    return 0;}

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