hdu 2829 斜率DP 板子

http://acm.hdu.edu.cn/showproblem.php?pid=2829
题意：

在一条直线型的铁路上，每个站点有各自的权重num[i]，每一段铁路（边）的权重（题目上说是战略价值什么的好像）是能经过这条边的所有站点的乘积之和.。然后给你m个炮弹，让你选择破坏掉m段铁路，使剩下的整条铁路的战略价值最小。

cost=∑i=lrVi(pre[r]−pre[i−1]−Vi)=∑i=lrVi(pre[r]−pre[i−1])−∑i=lrV2i

dp[i][j]表示到j为止的前面炸了i条路的得到最小总数

dp[i,j]=min(dp[i−1,k]+cost)=(pre[i]−pre[k])2−(pre2[i]−pre2[k])=

min(dp[i−1,k]+pre[i]2+pre[k]2−2∗pre[i]pre[k]−pre2[i]+pre2[k])

dp[i][j]=min(pre[i]2−pre2[i]−2pre[k]pre[i]+dp[i−1][k]+pre2[k]+pre[k]2)

代码中部分代码的解释

−2∗pr[k](k)

sq(pr[i])+pr2[i]+dp[i][−1](b)

pr[i](x)

#include <bits/stdc++.h>using namespace std;#define rep(i,a,b) for(int i=(a);i<=(b);i++)#define per(i,a,b) for(int i=(a);i>=(b);i--)#define pb push_back#define mp make_pair#define cl(a) memset((a),0,sizeof(a))#ifdef HandsomeHow#define dbg(x) cerr << #x << " = " << x << endl#else#define dbg(x)#endiftypedef long long ll;typedef unsigned long long ull;typedef pair <int, int> pii;const int inf=0x3f3f3f3f;const double eps=1e-8;const int mod=1000000007;const double pi=acos(-1.0);inline void gn(long long&x){    int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;}inline void gn(int&x){long long t;gn(t);x=t;}inline void gn(unsigned long long&x){long long t;gn(t);x=t;}ll gcd(ll a,ll b){return a? gcd(b%a,a):b;}ll powmod(ll a,ll x,ll mod){ll t=1ll;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}// (づ°ω°)づe★------------------------------------------------const int maxn = 1005;ll dp[maxn][maxn];typedef pair <ll, ll> Line;#define k first#define b secondstruct ConvexHull{    Line stk[maxn];    int l, r;    void init(){ l = r = 0; }    //sz r - l  [l,r)    bool cover(Line &a, Line &b, Line &c)    {        // line a and b cover line c        return (a.k - b.k) * (b.b - c.b) >= (b.b - a.b) * (c.k - b.k);        //  >= 下凸包 求最小值        //  <= 上凸包 求最大值    }    void add(Line p)    {        while(r - l > 1 && cover(p,stk[r-2],stk[r-1]))r--;        stk[r++] = p;    }    ll calc(ll x, Line l){ return l.k * x + l.b; }    ll ask(ll x)    {        while(r - l > 1 && calc(x,stk[l+1]) <= calc(x,stk[l]))l++;        return calc(x,stk[l]);    }}hull;#undef k#undef bll sq(ll x){ return x*x; }ll pr[maxn], pr2[maxn];ll v[maxn];int n,m;int main(){#ifdef HandsomeHow    freopen("data.in","r",stdin);    //freopen("data.out","w",stdout);#endif    while(true){        gn(n);gn(m);        if(n==0&&m==0)break;        rep(i,1,n){            gn(v[i]);            pr[i]=pr[i-1]+v[i];            pr2[i]=pr2[i-1]+sq(v[i]);        }        dp[0][0]=0;        rep(i,1,n)dp[i][0]=sq(pr[i])-pr2[i];        rep(_,1,m){            hull.init();            hull.add(mp(0,0));            rep(i,1,n){                if(i  <= _ + 1)                    dp[i][_] = 0;                else                    dp[i][_] = hull.ask(pr[i]) - pr2[i] + sq(pr[i]);                hull.add( mp( -2*pr[i],sq(pr[i]) + pr2[i] + dp[i][_-1] ) );            }        }        printf("%lld\n",dp[n][m]/2);    }    return 0;}