数形结合 + 二分凸壳3题
最近遇到了三道数形结合的题目，不同的动机都直接指向了凸包（凸壳），利用凸壳上斜率（极角）的单调性进行二分。
1 .一个在傻X那里淘到的一道数据结构题，from spoj：
维护一个数据结构，支持：序列区间加/减一个数， 求区间最大前缀和。
前面的部分是利用块状数组平衡复杂度， 最后一步需要维护：
max（s[i] + bj * i）;
这里构造所有的点（i，s[ i ] ), 对这些点求一个凸包（一个上凸壳，一个下凸壳）， 当bj为负时，在s的凸壳上二分（利用叉积），找到变化率刚好无法抵消bj的时候，bj为正时，同理。
# include <cstdlib>
# include <cstdio>
# include <cmath>
//# define int long long
using namespace std;
const long long oo = floor(1e18);
const int maxs = 1000, maxn = 100000+5;
int size, num, n, m, c, d, sl, sr;
long long begin[maxs], end[maxs], bj[maxs], tot[maxs], lf[maxs], rf[maxs], lz[maxs], rz[maxs];
long long stz[maxn], stf[maxn], pr[maxn], id[maxn], a[maxn];
inline long long max(long long x, long long y)
{
return x > y? x: y;
}
inline long long min(long long x, long long y)
{
return x < y? x: y;
}
inline long long abs1(long long x)
{
return x > 0? x: -x;
}
void change(int x)
{
int i;
for (i = begin[x]; i <= end[x]; i++) a[i] += bj[x];
for (tot[x] = 0, i = begin[x]; i <= end[x]; i++) tot[x] += a[i];
for (pr[begin[x]] = a[begin[x]], i = begin[x]+1; i <= end[x]; i++) pr[i] = pr[i-1] + a[i];
bj[x] = 0;
for (i = lz[x]; i <=rz[x]; stz[i++] = 0);
rz[x] = lz[x];
stz[lz[x]] = begin[x];
for (i = begin[x]+1; i <= end[x]; i++)
{
for (;(rz[x]>=lz[x] && pr[stz[rz[x]]] <= pr[i]) ||
(rz[x]>lz[x] && (abs1(pr[stz[rz[x]]] - pr[stz[rz[x]-1]])) * (i-stz[rz[x]])
>= (abs1(pr[i]-pr[stz[rz[x]]])) * (stz[rz[x]]-stz[rz[x]-1]));
rz[x]--);
stz[++rz[x]] = i;
}
for (i = lf[x]; i <=rf[x]; stf[i++] = 0);
rf[x] = lf[x];
stf[lf[x]] = begin[x];
for (i = begin[x]+1; i <= end[x]; i++)
{
for (;(rf[x]>lf[x] &&
(pr[stf[rf[x]]] - pr[stf[rf[x]-1]]) *(i - stf[rf[x]]) <= (pr[i] - pr[stf[rf[x]]]) * (stf[rf[x]] - stf[rf[x]-1]))
;
rf[x]--);
if (pr[stf[rf[x]]] < pr[i]) stf[++rf[x]] = i;
}
}
long long askmax(int x)
{
if (bj[x] >= 0)
{
long long ask = pr[stz[lz[x]]] + bj[x] * (stz[lz[x]] - begin[x] + 1);
int mid, ll = lz[x]+1, rr = rz[x];
if (ll > rr) return ask;
for (;ll < rr;)
{
mid = (ll + rr + 1) >> 1;
if (abs1(pr[stz[mid]] - pr[stz[mid-1]]) >= bj[x] * (stz[mid] - (stz[mid-1]) )) rr = mid - 1;
else ll = mid;
}
return max(ask, pr[stz[ll]]+bj[x] * (stz[ll] - begin[x]+1));
}
else
{
long long ask = pr[stf[lf[x]]] + bj[x] * (stf[lf[x]] - begin[x] + 1);
int mid, ll = lf[x]+1; int rr = rf[x];
rr = rf[x];
if (ll > rr) return ask;
for (;ll < rr;)
{
mid = (ll + rr + 1) >> 1;
if (pr[stf[mid]] - pr[stf[mid-1]] > abs(bj[x]) * (stf[mid] - (stf[mid-1]) )) ll = mid;
else rr = mid-1;
}
return max(ask, pr[stf[ll]]+bj[x] * (stf[ll]-begin[x]+1));
}
}
void read()
{
int i, j;
scanf("%d%d", &n, &m);
for (i = 1; i <=n; i++) scanf("%I64d", &a[i]);
size = floor(sqrt(n))+1;
for (i = 1, num = 1; i <=n; i+=size, num++)
{
lz[num]=lf[num] = begin[num] = i, end[num] = i+size-1;
if (i + size > n) end[num] = n;
for (j = begin[num]; j <= end[num]; j++)
id[j] = num, tot[num] += a[j];
}
for (i = 1; i <=num; i++)
change(i);
}
void modify(int l,int r, int d)
{
int i;
int gl = id[l], gr = id[r];
if (gr == gl)
{
for (i = l; i <= r; i++) a[i] += d;
change(gl);
}
else
{
for (i = l; i <= end[gl]; i++) a[i] += d;
for (i = begin[gr]; i <= r; i++) a[i] += d;
change(gl); change(gr);
for (i = gl+1; i < gr; i++) bj[i] += d;
}
}
long long query(int l, int r)
{
long long now = 0, ask = -oo;
int i, gl = id[l], gr = id[r];
for (i = 1; i < gl; now += tot[i]+ bj[i] * size, i++);
for (i = begin[gl]; i < l; now += a[i] + bj[gl], i++);
if (gl == gr)
{
for (i = l; i <= r; i++)
{
now += a[i]+bj[gl];
ask = max(ask, now);
}
}
else
{
for (i = l; i <= end[gl]; i++)
{
now += a[i] + bj[gl];
ask = max(ask, now);
}
for (i = gl+1; i < gr; i++)
{
ask = max(ask, now + askmax(i));
now += tot[i] + size * bj[i];
}
for (i = begin[gr]; i <= r; i++)
{
now += a[i] + bj[gr];
ask = max(ask, now);
}
}
return ask;
}
int main()
{
int i;
freopen("notdiff.in", "r", stdin);
freopen("notdiff.out", "w", stdout);
read();
for (i = 1; i <= m; i++)
{
scanf("%d", &c);
if (c == 1)
{
scanf("%d%d%d", &sl, &sr, &d);
modify(sl, sr, d);
}
else
{
scanf("%d%d", &sl, &sr);
printf("%I64d\n", query(sl, sr));
}
}
return 0;
}
2.wc2012 卓亮ppt / coci 2011&2012 contest3
一个工厂制造产品，有N个流程。第i个流程的时间系数是Ti。
有M个产品要制造，第i个产品的容易程度是Fi。一个产品j，在
流程i所需时间为TiFj。流程顺序不可颠倒，产品也必
须按给定的顺序制作。一旦一个流程完成，就交给下一个流程。
此时，下一个流程必须是空闲的，不然就会出错。问完成所有产
品需要的时间。
数据满足1 ≤ N ≤ 100000，1 ≤ M ≤ 100000。
贪心的认为，每个产品尽量早的开始生产；
令time[i]为i产品的开始时间，则for (all j) time[i] + Fi*sigma(Tj) >= time[i+1] +Fi+1 *sigma(Tj-1);
令Si = sigma（Ti）；
则time[i +1] - time[i] >= Fi*Sj - Fi+1* Sj-1； 后面就是叉积~\(≧▽≦)/~啦啦啦
如果求出最大叉积，就可以有time[i] 推出time[j]；
后面那个式子，就是对于一个（Fi, Fi+1） 求出对于所有的（Sj-1， Sj）中叉积最大且为正的那个。
那么对于（Sj-1, Sj）维护一个上凸壳，二分一下就可以了 。
# include <cstdlib>
# include <cmath>
# include <cstdio>
# include <cstring>
using namespace std;
const int maxn = 200000;
struct point
{
long long x, y;
}sd[maxn];
int que[maxn];
int n, m, i;
long long s[maxn];
int e[maxn], t[maxn];
long long cross(point a, point b, point c, point d)
{
long long x1=b.x-a.x,y1=b.y-a.y,x2=d.x-c.x,y2=d.y-c.y;
return x1*y2-x2*y1;
}
void prepare()
{
for (i = 1; i < n; i++)
{
for (;que[0]>1&cross(sd[que[que[0]-1]], sd[i], sd[que[que[0]-1]], sd[que[que[0]]])<=0; que[0]--);
que[++que[0]]= i;
}
/* cut = que[0];
for (i = n-1; i >= 1; i--)
{
for (;que[0]>1&cross(sd[que[0]-1], sd[i], sd[que[0]-1], sd[que[0]])<=0; que[0]--);
que[++que[0]]= i;
}
*/
}
inline long long max(long long x, long long y)
{
return x > y?x: y;
}
long long check(int x, int y)
{
point c; c.x=x; c.y=y;
long long ask = cross(sd[0], c, sd[0], sd[que[que[0]]]);
int mid, l = 1, r = que[0]-1;
for (;l <= r;)
{
mid = l+r>>1;
ask = max(ask, cross(sd[0], c, sd[0], sd[que[mid]]));
if (cross(sd[que[mid]], sd[que[mid+1]], sd[0], c)<0) l = mid+1; else r = mid-1;
}
return ask;
}
int main()
{
freopen("traka.in", "r", stdin);
freopen("traka.out", "w", stdout);
scanf("%d%d", &n, &m);
for (i = 1; i <= n; i++) scanf("%d", &t[i]);
for (i = 1; i <= m; i++) scanf("%d", &e[i]);
for (i = 1; i <= n; i++) s[i] = s[i-1] + t[i];
for (i = 1; i < n; i++) sd[i].x = s[i], sd[i].y = s[i+1];
prepare();
long long btime = 0, inc;
for (i = 1; i < m; i++)
{
inc = max(1LL*e[i]*t[1], check(e[i],e[i+1]));
btime += inc;
}
btime += s[n]* e[m];
printf("%I64d", btime);
return 0;
}
还是wc2012卓亮论文中提到的：
某人要组织一场比赛。她有N道备选题，这场比赛有K题。每位选手要做这K题。
她想，如果选手把这K题全做出来，选手会觉得这个比赛过于简单，很无趣。
但如果选手只做出了很少的题目，又会觉得很难过。
因此她想选这K道题，使得解出恰好K − 1题的概率尽量大。假设她已经进行了实验，得出了每道题被解出的概率。1 ≤ K ≤ N ≤ 36。
令ai是选出的第i题做出的概率，
那么题目是求max（(1−a1)a2a3...aK+a1(1−a2)a3a4...aK+···+a1a2...aK−1(1−aK)）；
化一化,变成pai ai * sigma（（1-a[i])/a[i]）, 直接裸搜是不行的，考虑折半搜索，那么左式分成两部分，会惊喜的发现是一个叉积的形式，求叉积的最大值，那么先搜左边一半，建立凸壳，再搜右边一半，二分即可。
不能输公式所以写不清楚，卓亮的论文上倒是很清楚。
# include <cstdlib>
# include <cstdio>
# include <cmath>
# include <cstring>
using namespace std;
const int maxn = 30;
const double eps=1e-10;
struct point
{
double x,y;
};
int have[maxn];
int n, cut, k;
double ans, a[maxn*2];
inline bool bezero(double x)
{
return x<eps&&x>-eps?true:false;
}
inline double cross(point a, point b, point c, point d)
{
double x1=b.x-a.x,y1=b.y-a.y,x2=d.x-c.x,y2=d.y-c.y;
return x1*y2-x2*y1;
}
inline double cj(point a, point b)
{
return a.x*b.y-a.y*b.x;
}
inline double max(double x, double y)
{
return x>y?x:y;
}
struct forkcase
{
int que[100000]; point sd[100000];
void sort(int l, int r)
{
int i = l, j = r; point d = sd[l+r>>1], tmp;
for (;i <= j;)
{
for (;d.x-sd[i].x> eps||(bezero(d.x-sd[i].x) && d.y-sd[i].y> eps);i++);
for (;d.x-sd[j].x<-eps||(bezero(d.x-sd[j].x) && d.y-sd[j].y<-eps);j--);
if (i <= j)
tmp=sd[i], sd[i]=sd[j], sd[j]=tmp,i++,j--;
}
if (i< r) sort(i, r);
if (l< j) sort(l, j);
}
void tidy(int p)
{
int i;
que[0] = 1; que[1] = 1;
for (i = 2; i <= have[p]; i++)
{
for (;que[0]>1 && cross(sd[que[que[0]-1]], sd[que[que[0]]], sd[que[que[0]-1]],sd[i])<=eps; que[0]--);
que[++que[0]] = i;
}
}
void updata(point g, int p)
{
point d=(point){0,0};
if (p < 0) return;
ans = max(ans,
cj(sd[que[que[0]]], g));
int l= 1, r= que[0]-1, mid;
for (;l <= r;)
{
mid = l+r>>1; ans=max(ans, cj(sd[que[mid]],g));
if (cross(d, g, sd[que[mid]], sd[que[mid+1]])<0)
l = mid+1;
else r = mid-1;
}
}
}kcase[maxn];
void dfs1(int now, int tot, double s, double t)
{
int i; if (tot > k) return;
if (tot == k) ans = max(ans, s*t);
have[tot]++;kcase[tot].sd[have[tot]]=(point){t*s, -t};
for (i = now; i <= cut; i++)
dfs1(i+1, tot+1, s+(1-a[i])/a[i], t*a[i]);
}
void dfs2(int now, int tot, double s, double t)
{
if (tot > k) return;
if (tot== k) ans = max(ans, s*t);
int i; point g = (point){t*s, t}; kcase[k-tot].updata(g, k-tot);
for (i = now; i <= n; i++)
dfs2(i+1, tot+1, s+(1-a[i])/a[i], t*a[i]);
}
int main()
{
int i;
freopen("pro.in", "r", stdin);
freopen("pro.out", "w", stdout);
scanf("%d%d", &n, &k);
for (i = 1; i <= n; i++) scanf("%lf", &a[i]);
for (i = 1; i <= n; i++) a[i]/=100;
cut = n/2;
dfs1(1,0,0,1);
for (i = 0; i <= k; i++)
kcase[i].sort(1, have[i]);
for (i = 0; i <= k; i++)
kcase[i].tidy(i);
ans = 0;
dfs2(cut+1,0,0,1);
printf("%.4lf", ans);
return 0;
}