bzoj 1096: [ZJOI2007]仓库建设

同样是一个斜率优化，设f[i]表示在i处建仓库，f[i] = f[j] + cal(j,i) + c[i];

一开始cal想了我好久，一直只想到o（n）cal。。。

后面看着花花想cal的实现，一下子就想出来了！！！

斜率优化的一般方法应该是 f[i] + 只与i有关的看作c，只与j有关的看作by，与ij有关的j看作x，i看作a，再用向量积去做

等下再用决策单调性优化写下这道题。。

妈蛋一开始队列写错了卡了15分钟！！！最近老是犯些SB错误。。还是静不下来啊

/*
ID:WULALA
PROB:bzoj1096_slope
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <ctime>
#include <set>
#define N 1000008 
#define M
#define mod
#define mid(l,r) ((l+r) >> 1)
#define INF 0x7ffffff
using namespace std;

long long l,r,n,p[N],que[N],f[N],d[N],c[N],s[N],sum[N];

void init()
{
    scanf("%lld",&n);
    for (int i = 1;i <= n;i++)
    {
        scanf("%lld%lld%lld",&d[i],&p[i],&c[i]);
        s[i] = s[i-1] + sum[i-1] * (d[i] - d[i-1]);
        sum[i] = sum[i-1] + p[i];
    }
    l = r = 1;
}

long long cal(long long x,long long y)
{
    long long r = (f[x] + s[y] - s[x] - sum[x] * (d[y] - d[x]) + c[y]);
    return r;
}

long long cross(long long o,long long a,long long b)
{
    long long xo = sum[o],yo = f[o] - s[o] + sum[o] * d[o];
    long long xa = sum[a],ya = f[a] - s[a] + sum[a] * d[a];
    long long xb = sum[b],yb = f[b] - s[b] + sum[b] * d[b];
    xa -= xo; ya -= yo;
    xb -= xo; yb -= yo;
    return (xa * yb - xb * ya);
}

void debug(long long a)
{
    printf("%lld %lld
",que[l],f[a]);
}

int main()
{
    init();
    for (int i = 1;i <= n;i++)
    {
        while (l < r && cal(que[l],i) > cal(que[l+1],i)) 
            l++;
        f[i] = cal(que[l],i);
//        debug(i);
        while (l < r && cross(que[r-1],que[r],i) < 0) r--;
        que[++r] = i;
    }
    printf("%lld
",f[n]);
    return 0;
}

 决策单调性

/*
ID:WULALA
PROB:bzoj1096_mono
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <ctime>
#include <set>
#define N 1000008 
#define M
#define mod
#define mid(l,r) ((l+r) >> 1)
#define INF 0x7ffffff
using namespace std;

long long l,r,n,p[N],que[N],f[N],d[N],c[N],s[N],sum[N],st[N];

void init()
{
    scanf("%lld",&n);
    for (int i = 1;i <= n;i++)
    {
        scanf("%lld%lld%lld",&d[i],&p[i],&c[i]);
        s[i] = s[i-1] + sum[i-1] * (d[i] - d[i-1]);
        sum[i] = sum[i-1] + p[i];
    }
    for (int i = 1;i <= n+1;i++) st[i] = n + 1;
    l = r = 1; st[0] = 1;
}

long long cal(long long x,long long y)
{
    long long r = (f[x] + s[y] - s[x] - sum[x] * (d[y] - d[x]) + c[y]);
    return r;
}

long long find(long long a)
{
    long long top = r + 1,bot = l;//top应= r + 1;!! 
    while (bot != top)
    {
        if (st[que[mid(bot,top)]] <= a) bot = mid(bot,top) + 1;
        else top = mid(bot,top);
    }
    return (bot - 1);
}

int main()
{
    init();
    for (int i = 1;i <= n;i++)
    {
        f[i] = cal(que[find(i)],i); 
        while (r && cal(que[r],st[que[r]]) > cal(i,st[que[r]])) r--; 
        int bot = st[que[r]],top = n + 1;
        while (bot != top)
        {
            if (cal(que[r],mid(top,bot)) > cal(i,mid(top,bot))) top = mid(top,bot);
            else bot = mid(top,bot) + 1;
        }
        if (bot == n + 1) continue;
        que[++r] = i;
        st[i] = bot;
    }
    printf("%lld
",f[n]);
    return 0;
}

感觉斜率优化式子写出来以后比决策单调性还好写些（>__<）