Uva 12436 Rip Van Winkle's Code

Rip Van Winkle was fed up with everything except programming. One day he found a problem whichrequired to perform three types of update operations (A, B, C), and one query operation S over an arraydata[]. Initially all elements of data are equal to 0. Though
Rip Van Winkle is going to sleep for 20years, and his code is also super slow, you need to perform the same update operations and output theresult for the query operation S in an efficient way.

long long data[250001];

void A( int st, int nd ) {

for( int i = st; i \le nd; i++ ) data[i] = data[i] + (i - st + 1);

}

void B( int st, int nd ) {

for( int i = st; i \le nd; i++ ) data[i] = data[i] + (nd - i + 1);

}

void C( int st, int nd, int x ) {

for( int i = st; i \le nd; i++ ) data[i] = x;

}

long long S( int st, int nd ) {

long long res = 0;

for( int i = st; i \le nd; i++ ) res += data[i];

return res;

}

Input

The first line of input will contain T (≤ 4 ∗ 105) denoting the number of operations. Each of the nextT lines starts with a character (‘A’, ‘B’, ‘C’ or ‘S’), which indicates the type of operation. Character ‘A’,‘B’ or ‘S’ will be followed by two integers,
st and nd in the same line. Character ‘C’ is followed by threeintegers, st, nd and x. It’s assumed that, 1 ≤ st ≤ nd ≤ 250000 and −105 ≤ x ≤ 105. The meaningsof these integers are explained by the code of Rip Van Winkle.

Output

For each line starting with the character ‘S’, print S(st, nd) as defined in the code.

Sample Input

7

A 1 4

B 2 3

S 1 3

C 3 4 -2

S 2 4

B 1 3

S 2 4

Sample Output

9

0

3

这题是区间更新，这题比較麻烦，做了非常长时间。先用线段树维护l,r,add1（线段左端点加的值）,add2（线段右端点加的值）,step（区间的公差，右边减去左边的）,sum（区间总和）,flag（推断区间是否数字同样）,value（区间数字都同样时的数值大小）.我的思路是每一次更新，都把这一段的sum值直接表示出来。假设更新的这条线段小于当前线段，那么先不更新sum值。而是b[th].sum=b[lth].sum+b[rth].sum;

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define lth th<<1
#define rth th<<1|1
#define inf 99999999
#define pi acos(-1.0)
#define MOD 100000007
#define maxn 250050
struct node{
    int l,r;
    ll value,flag;  //flag表示这段是不是值都是同样的,value是这段的值
    ll add1,step,add2; //add1表示左端点加的值，add2表示右端点，step表示这段的公差
    ll sum;
}b[4*maxn];

void build(int l,int r,int th)
{
    int mid;
    b[th].l=l;b[th].r=r;
    b[th].value=0;b[th].flag=1;
    b[th].add1=b[th].step=b[th].add2=0;
    b[th].sum=0;
    if(l==r)return;
    mid=(l+r)/2;
    build(l,mid,lth);
    build(mid+1,r,rth);
}
void pushdown(int th)
{
    int mid;
    mid=(b[th].l+b[th].r)/2;
    if(b[th].flag){
        b[th].flag=0;
        b[lth].flag=1;
        b[lth].value=b[th].value;
        b[lth].add1=b[lth].add2=b[lth].step=0;
        b[lth].sum=(b[lth].r-b[lth].l+1)*b[th].value;

        b[rth].flag=1;
        b[rth].value=b[th].value;
        b[rth].add1=b[rth].add2=b[rth].step=0;
        b[rth].sum=(b[rth].r-b[rth].l+1)*b[th].value;
    }

    ll add1,add2;
    add1=b[th].add1; add2=b[th].add1+(mid-b[th].l)*b[th].step;
    b[lth].add1+=add1;
    b[lth].add2+=add2;
    b[lth].step+=b[th].step;
    b[lth].sum+=(add1+add2)*(b[lth].r-b[lth].l+1)/2;

    ll add3,add4;
    add3=add2+b[th].step;add4=add3+(b[th].r-(mid+1))*b[th].step;
    b[rth].add1+=add3;
    b[rth].add2+=add4;
    b[rth].step+=b[th].step;
    b[rth].sum+=(add3+add4)*(b[rth].r-b[rth].l+1)/2;

    b[th].add1=b[th].add2=b[th].step=0;
}
void pushup(int th)
{
    b[th].sum=b[lth].sum+b[rth].sum;
}

void update(int l,int r,ll add,int f,int th)
{
    int mid;
    if(b[th].l==l && b[th].r==r){
        if(f==1){
            b[th].add1+=add;
            b[th].add2+=add+b[th].r-b[th].l;
            b[th].step+=1;
            b[th].sum+=(add+add+b[th].r-b[th].l)*(b[th].r-b[th].l+1)/2;
            return;
        }
        else if(f==2){
            b[th].add1+=add+b[th].r-b[th].l;
            b[th].add2+=add;
            b[th].step-=1;
            b[th].sum+=(add+add+b[th].r-b[th].l)*(b[th].r-b[th].l+1)/2;
            return;
        }
        else if(f==3){
            b[th].flag=1;
            b[th].value=add;
            b[th].sum=b[th].value*(b[th].r-b[th].l+1);
            b[th].add1=b[th].add2=b[th].step=0;
            return;
        }
    }
    pushdown(th);
    mid=(b[th].l+b[th].r)/2;
    if(r<=mid)update(l,r,add,f,lth);
    else if(l>mid)update(l,r,add,f,rth);
    else{
        if(f==1){
            update(l,mid,add,f,lth);
            update(mid+1,r,add+(mid+1-l),f,rth);
        }
        else if(f==2){
            update(l,mid,add+(r-mid),f,lth);
            update(mid+1,r,add,f,rth);
        }
        else if(f==3){
            update(l,mid,add,f,lth);
            update(mid+1,r,add,f,rth);
        }
    }
    pushup(th);
}
ll question(int l,int r,int th)
{
    int mid;
    if(b[th].l==l && b[th].r==r){
        return b[th].sum;
    }
    pushdown(th);
    mid=(b[th].l+b[th].r)/2;
    if(r<=mid)return question(l,r,lth);
    else if(l>mid)return question(l,r,rth);
    else{
        return question(l,mid,lth)+question(mid+1,r,rth);
    }
}
int main()
{
    int m,i,j,T,c,d;
    ll n,num;
    char s[10];
    while(scanf("%lld",&n)!=EOF)
    {
        build(1,250010,1);
        for(i=1;i<=n;i++){
            scanf("%s%d%d",s,&c,&d);
            if(s[0]=='A'){
                update(c,d,1,1,1);
            }
            else if(s[0]=='B'){
                update(c,d,1,2,1);
            }
            else if(s[0]=='C'){
                scanf("%lld",&num);
                update(c,d,num,3,1);
            }
            else if(s[0]=='S'){
                printf("%lld\n",question(c,d,1) );
            }
        }
    }
    return 0;
}