Mango DS Training #48 ---线段树2 解题手记

Training address： http://acm.hust.edu.cn/vjudge/contest/view.action?cid=38966#overview

 

A.Count Color --- POJ 2777

这题初看不好下手，再想想，T<=30，这时想到颜色可以用二进制来表示，然后父节点颜色种类为子节点的按位或得出的结果中化为二进制时1的个数，然后就是无脑线段树了。。

 

代码：

#include <iostream>
#include <cstdio>
#include <utility>
#include <cstdlib>
using namespace std;
#define N 100010

struct node
{
    int mark;
    int sum;
}tree[4*N];

void pushup(int rt)
{
    tree[rt].sum = tree[2*rt].sum | tree[2*rt+1].sum;
}

void build(int l,int r,int rt)
{
    tree[rt].sum = 1;
    tree[rt].mark = 1;
    if(l == r)
        return;
    int mid = (l+r)/2;
    build(l,mid,2*rt);
    build(mid+1,r,2*rt+1);
    pushup(rt);
}

void pushdown(int l,int r,int rt)
{
    if(!tree[rt].mark)
        return;
    if(tree[rt].mark)
    {
        tree[2*rt].sum = tree[2*rt+1].sum = tree[rt].sum;
        tree[2*rt].mark = tree[2*rt+1].mark = tree[rt].mark;
        tree[rt].mark = 0;
    }
}

void make(int l,int r,int aa,int bb,int co,int rt)
{
    if(aa<=l && bb>=r)
    {
        tree[rt].mark = 1;
        tree[rt].sum = 1<<(co-1);
        return;
    }
    pushdown(l,r,rt);
    int mid = (l+r)/2;
    if(aa<=mid)
        make(l,mid,aa,bb,co,2*rt);
    if(bb>mid)
        make(mid+1,r,aa,bb,co,2*rt+1);
    pushup(rt);
}

int query(int l,int r,int aa,int bb,int rt)
{
    if(aa>r||bb<l)
        return 0;
    if(aa<=l&&bb>=r)
    {
        return tree[rt].sum;
    }
    pushdown(l,r,rt);
    int mid = (l+r)/2;
    return query(l,mid,aa,bb,2*rt)|query(mid+1,r,aa,bb,2*rt+1);
}

int main()
{
    int n,t,o;
    int i;
    int aa,bb,co;
    char ss[5];
    scanf("%d%d%d",&n,&t,&o);
    build(1,n,1);
    int cnt;
    for(i=0;i<o;i++)
    {
        scanf("%s",ss);
        if(ss[0] == 'C')
        {
            scanf("%d%d%d",&aa,&bb,&co);
            make(1,n,aa,bb,co,1);
        }
        else if(ss[0] == 'P')
        {
            scanf("%d%d",&aa,&bb);
            int res;
            if(aa<=bb)
                res = query(1,n,aa,bb,1);
            else
                res = query(1,n,bb,aa,1);
            cnt = 0;
            while(res)
            {
                if(res&1)
                    cnt++;
                res>>=1;
            }
            printf("%d\n",cnt);
        }
    }
    return 0;
}

 

B.Who Gets the Most Candies --- POJ 2886

(题解借鉴： ahfywff）

    本题利用反素数的概念。反素数的定义：对于任何正整数x，其约数的个数记做f(x)。例如f(1)=1,f(6)=4。如果某个正整数x满足：对于任意i(0<i<x)，都有f(i)<f(x)，则称x为反素数。对于本题，设pos为不大于N的反素数，则第pos个出圈的孩子得到的糖果最多，为pos的约数个数。

　　 出圈过程有点类似约瑟夫环。假设当前出圈的是剩余孩子中的第K个，他手中的数字为A。

    若A大于零，下一个出圈的就应该是剩余孩子中的第(K-1+A-1)%n+1个；

    若A小于零，下一个出圈的就应该是剩余孩子中的第((K-1+A)%n+n)%n+1个。

    问题的关键是如何求得出圈孩子的原始位置，线段树的每个节点的sum存储了所在区间还有多少孩子留下，查询节点rt中第num个孩子的原始位置时，如果num<=st[2*rt].sum，则在左孩子节点中查询第num个孩子的原始位置；否则在右孩子节点中查询第num-st[2*rt].sum个孩子的原始位置。

 

  代码：

/*12152 KB    1079 ms*/
#include <iostream>
#include <cstdio>
using namespace std;
#define N 500010

int anti[37]={1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,20160,25200,27720,45360,50400,
           55440,83160,110880,166320,221760,277200,332640,498960,500001};
int factor[37]={1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60,64,72,80,84,90,96,100,108,120,128,144,160,168,180,192,200,1314521};

int tree[4*N];
char name[N][11];
int num[N];

void build(int l,int r,int rt)
{
    tree[rt] = r-l+1;
    if(l == r)
        return;
    int mid = (l+r)/2;
    build(l,mid,2*rt);
    build(mid+1,r,2*rt+1);
}

int query(int l,int r,int pos,int rt)
{
    tree[rt]--;
    if(l == r)
        return l;
    int mid = (l+r)/2;
    if(pos<=tree[2*rt])
        return query(l,mid,pos,2*rt);
    else
        return query(mid+1,r,pos-tree[2*rt],2*rt+1);
}

int main()
{
    int n,k,pos,Maxcandy,i;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        i=0;
        int CN = n;
        while(anti[i]<=n)
            i++;
        pos = anti[i-1];
        Maxcandy = factor[i-1];
        build(1,n,1);
        for(i=1;i<=n;i++)
        {
            scanf("%s %d",name[i],&num[i]);
        }
        int flag;
        for(i=1;i<=pos;i++)
        {
            n--;
            flag = query(1,CN,k,1);
            if(n==0)
                break;
            if(num[flag]>0)
                k = (k + num[flag] - 2)%n + 1;
            else
                k = ((k + num[flag] - 1)%n+n)%n + 1;
        }
        printf("%s %d\n",name[flag],Maxcandy);
    }
}

 

G.Fast Matrix Operations ---UVA 11992

  这题其实也不太难搞，关键是各方面要维护到，我写了两个多小时啊，最后还是有些地方没有照顾到，哎，太弱咯。。感觉自己在维护值方面还差一点火候。 这题看行数不超过20，想到在每一行都建一颗线段树，然后就搞吧。。代码有点长，将就着看吧。

(A.M : Attention to Maintain)

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <utility>
#include <cstdlib>
#define Mod 1000000007
using namespace std;
#define N 1000010

struct node
{
    int mini,maxi,sum;
    int addmark,setmark;
}tree[21][4*N];

int i;

void pushup(int i,int rt)
{
    tree[i][rt].sum = tree[i][2*rt].sum + tree[i][2*rt+1].sum;
    tree[i][rt].mini = min(tree[i][2*rt].mini,tree[i][2*rt+1].mini);
    tree[i][rt].maxi = max(tree[i][2*rt].maxi,tree[i][2*rt+1].maxi);
}

void build(int i,int l,int r,int rt)
{
    tree[i][rt].sum = tree[i][rt].mini = tree[i][rt].maxi = 0;
    tree[i][rt].setmark = -1;
    tree[i][rt].addmark = 0;
    if(l == r)
        return;
    int mid = (l+r)/2;
    build(i,l,mid,2*rt);
    build(i,mid+1,r,2*rt+1);
    pushup(i,rt);
}

void pushdown(int i,int l,int r,int rt)
{
    if(tree[i][rt].setmark == -1 && tree[i][rt].addmark == 0)
        return;
    int mid = (l+r)/2;
    if(tree[i][rt].setmark >= 0)
    {
        tree[i][2*rt].sum = tree[i][rt].setmark*(mid-l+1);
        tree[i][2*rt+1].sum = tree[i][rt].setmark*(r-mid);
        tree[i][2*rt].mini = tree[i][2*rt+1].mini = tree[i][rt].setmark;  //A.M 
        tree[i][2*rt].maxi = tree[i][2*rt+1].maxi = tree[i][rt].setmark;  //A.M
        tree[i][2*rt].addmark = tree[i][2*rt+1].addmark = 0;  // 这个要写 
        tree[i][2*rt].setmark = tree[i][2*rt+1].setmark = tree[i][rt].setmark;
        tree[i][rt].setmark = -1;
    }
    if(tree[i][rt].addmark > 0)
    {
        tree[i][2*rt].sum += tree[i][rt].addmark*(mid-l+1);
        tree[i][2*rt+1].sum += tree[i][rt].addmark*(r-mid);
        tree[i][2*rt].maxi += tree[i][rt].addmark;  //A.M
        tree[i][2*rt].mini += tree[i][rt].addmark;  //A.M
        tree[i][2*rt+1].maxi += tree[i][rt].addmark;  //A.M
        tree[i][2*rt+1].mini += tree[i][rt].addmark;  //A.M
        tree[i][2*rt].addmark += tree[i][rt].addmark;
        tree[i][2*rt+1].addmark += tree[i][rt].addmark;
        tree[i][rt].addmark = 0;
    }
}

void add(int l,int r,int aa,int bb,int val,int rt)
{
    if(aa>r||bb<l)
        return;
    if(aa<=l&&bb>=r)
    {
        tree[i][rt].addmark += val;
                                   //tree[i][rt].setmark = -1;  --不要写这个 
        tree[i][rt].sum += (r-l+1)*val;
        tree[i][rt].maxi += val;
        tree[i][rt].mini += val;
        return;
    }
    pushdown(i,l,r,rt);
    int mid = (l+r)/2;
    if(aa<=mid)
        add(l,mid,aa,bb,val,2*rt);
    if(bb>mid)
        add(mid+1,r,aa,bb,val,2*rt+1);
    pushup(i,rt);
}

void setval(int l,int r,int aa,int bb,int val,int rt)
{
    if(aa>r||bb<l)
        return;
    if(aa<=l&&bb>=r)
    {
        tree[i][rt].setmark = val;
        tree[i][rt].addmark = 0;
        tree[i][rt].sum = val*(r-l+1);
        tree[i][rt].maxi = tree[i][rt].mini = val;
        return;
    }
    pushdown(i,l,r,rt);
    int mid = (l+r)/2;
    if(aa<=mid)
        setval(l,mid,aa,bb,val,2*rt);
    if(bb>mid)
        setval(mid+1,r,aa,bb,val,2*rt+1);
    pushup(i,rt);
}

struct node_ans
{
    int sum;
    int mini,maxi;
};

node_ans query(int l,int r,int aa,int bb,int rt)
{
    node_ans res,ka1,ka2;
    if(aa<=l && bb>=r)
    {
        res.sum = tree[i][rt].sum;
        res.maxi = tree[i][rt].maxi;
        res.mini = tree[i][rt].mini;
        return res;
    }
    pushdown(i,l,r,rt);
    int mid = (l+r)/2;
    if(bb<=mid)
        return query(l,mid,aa,bb,2*rt);
    else if(aa>mid)
        return query(mid+1,r,aa,bb,2*rt+1);
    else
    {
        ka1 = query(l,mid,aa,bb,2*rt);
        ka2 = query(mid+1,r,aa,bb,2*rt+1);
        res.sum = ka1.sum + ka2.sum;
        res.maxi = max(ka1.maxi,ka2.maxi);
        res.mini = min(ka1.mini,ka2.mini);
        return res;
    }
}

int main()
{
    int r,c,m;
    int k,zuo;
    int x1,y1,x2,y2,val;
    int sum,mmax,mmin;
    while(scanf("%d%d%d",&r,&c,&m)!=EOF)
    {
        for(i=1;i<=r;i++)
        {
            build(i,1,c,1);
        }
        for(k=0;k<m;k++)
        {
            scanf("%d",&zuo);
            if(zuo == 1)
            {
                scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&val);
                for(i=x1;i<=x2;i++)
                {
                    add(1,c,y1,y2,val,1);
                }
            }
            else if(zuo == 2)
            {
                scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&val);
                for(i=x1;i<=x2;i++)
                {
                    setval(1,c,y1,y2,val,1);
                }
            }
            else 
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                node_ans la;
                sum = 0;
                mmax = -Mod;
                mmin = Mod;
                for(i=x1;i<=x2;i++)
                {
                    la = query(1,c,y1,y2,1);
                    sum += la.sum;
                    mmax = max(mmax,la.maxi);
                    mmin = min(mmin,la.mini);
                }
                printf("%d %d %d\n",sum,mmin,mmax);
            }
        }
    }
}