bzoj 1010: [HNOI2008]玩具装箱toy

这道题是一道很简单的DP，方程很容易列出来，

只不过要优化，因为n^2过不了。

膜拜了一下搞神的《浅析1D1D动态规划的优化(zzx)》

既可以用斜率优化也可以用单调性优化。

单调性证明就不说了。

斜率优化还不太懂所以只写了单调性优化

/*
ID:WULALA
PROB:bzoj1010
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <ctime>
#include <set>
#define N 50008
#define M
#define mod
#define mid(l,r) ((l+r) >> 1)
#define INF 0x7ffffff
using namespace std;

int n,L,que[N],st[N],l = 1,r = 1;
long long f[N],pre[N];
inline long long sqr(long long a)
{
    return (a*a);
}

inline long long cal(int s,int t)
{
    return (f[s] + sqr(t - s  - 1 + pre[t] - pre[s] - L));
}

int find(int a)
{
    int bot = 1,top = r + 1;
    while (bot != top)
    {
        if (st[que[mid(bot,top)]] > a) top = mid(bot,top);
        else bot = mid(bot,top) + 1;
    }//寻找第一个比a小可以先找第一个比a大 
    return (bot - 1);
}

int work()
{
    memset(que,255,sizeof(que));
    que[1] = 0;
    st[0] = 1; 
    for (int i = 1;i <= n+1;i++) st[i] = n + 1;//方便查找
    for (int i = 1;i <= n;i++)
    {
        f[i] = cal(que[find(i)],i);
        while (st[que[r]] > i && r && cal(i,st[que[r]]) < cal(que[r],st[que[r]])) r--;
        if (!r)
        {
            que[++r] = i;
            st[i] = i;
            continue ;
        }
        int bot = st[que[r]],top = n + 1;
        while (bot != top)
        {
            if (cal(que[r],mid(bot,top)) >= cal(i,mid(bot,top))) top = mid(bot,top);
            else bot = mid(bot,top) + 1;
        }
        if (bot == n + 1) continue ;//此决策无法入栈
        que[++r] = i;
        st[i] = bot;
    }
}

int main()
{
    scanf("%d%d",&n,&L);
    for (int i = 1;i <= n;i++)
    {
        scanf("%d",&pre[i]);
        pre[i] += pre[i-1];
    }
    work();
    printf("%lld",f[n]);
    return 0;
}

 昨天和洛神聊了下天，把斜率优化学了，顺手写了一下

/*
ID:WULALA
PROB:bzoj1010 slope
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <ctime>
#include <set>
#define N 50008
#define M
#define mod
#define mid(l,r) ((l+r) >> 1)
#define INF 0x7ffffff
using namespace std;

int n,L,que[N],l,r;
long long pre[N],f[N];

inline long long sqr(long long a)
{
    return (a*a);
}

inline long long cal(int s,int t)
{
    return (f[s] + sqr(t - s  - 1 + pre[t] - pre[s] - L));
}

void init()
{
    scanf("%d%d",&n,&L);
    for (int i = 1;i <= n;i++)
    {
        scanf("%lld",&pre[i]);
        pre[i] += pre[i-1];
    }
    memset(que,255,sizeof(que));
    l = r = que[0] = 0;
}

long long det(int o,int a,int b)
{
    long long xo = pre[o],yo = f[o] + sqr(pre[o]) - 2 * l * pre[o]; 
    long long xa = pre[a],ya = f[a] + sqr(pre[a]) - 2 * l * pre[a]; 
    long long xb = pre[b],yb = f[b] + sqr(pre[b]) - 2 * l * pre[b];
    xa -= xo; xb -= xo;
    ya -= yo; yb -= yo;
    return (xa * yb - xb * ya); 
}

void debug()
{
    printf("\n");
    for (int i = l;i <= r;i++) printf("%d ",que[i]);
    printf("\n");
    for (int i = 1;i <= n;i++) printf("%d ",f[i]);
    printf("\n");
}

void work()
{
    for (int i = 1;i <= n;i++)
    {
        while (cal(que[l],i) >= cal(que[l+1],i) && que[l+1] != -1) l++;
        f[i] = cal(que[l],i);
        if (f[i] < 0) printf("error: i");
        while (det(que[r-1],que[r],i) < 0 && r > l) r--;
        que[++r] = i;
//        debug(); 
    }
}

int main()
{
    init();
    work();
    printf("%lld\n",f[n]);
    return 0;
}