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 longusing 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;}