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
7A 1 4B 2 3S 1 3C 3 4 -2S 2 4B 1 3S 2 4
Sample Output
9
0
3
这题是区间更新，，这题比较麻烦，做了很长时间。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（判断区间是否数字相同）,value（区间数字都相同时的数值大小）.
#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<algorithm>using namespace std;#define maxn 250060#define ll long longll sum;struct node{ll l,r,add1,add2,step,sum,flag,v; //全改成long long }b[8*maxn];void build(ll l,ll r,ll i){ll mid;b[i].l=l;b[i].r=r;b[i].add1=b[i].add2=b[i].step=0;b[i].flag=0;b[i].sum=b[i].v=0;if(l==r)return;mid=(l+r)/2;build(l,mid,i*2);build(mid+1,r,i*2+1);}void pushdown(ll i){ll mid,add1,add2,add3,add4;mid=(b[i].l+b[i].r)/2;if(b[i].l==b[i].r)return;if(b[i].flag==0){b[i*2].flag=b[i*2+1].flag=0;b[i*2].v=b[i*2+1].v=b[i].v;b[i*2].add1=b[i*2].add2=b[i*2+1].add1=b[i*2+1].add2=0;b[i*2].step=b[i*2+1].step=0;b[i*2].sum=(b[i*2].r-b[i*2].l+1)*b[i*2].v;b[i*2+1].sum=(b[i*2+1].r-b[i*2+1].l+1)*b[i*2+1].v;b[i].flag=-1;}if(b[i].add1 || b[i].add2 || b[i].step){add1=b[i].add1;add2=b[i].add1+b[i].step*(mid-b[i].l);add3=add2+b[i].step;add4=b[i].add2;b[i*2].add1+=add1;b[i*2].add2+=add2;b[i*2+1].add1+=add3;b[i*2+1].add2+=add4;b[i*2].sum+=(b[i*2].r-b[i*2].l+1)*(add1+add2)/2;b[i*2+1].sum+=(b[i*2+1].r-b[i*2+1].l+1)*(add3+add4)/2;b[i*2].step+=b[i].step;b[i*2+1].step+=b[i].step;b[i].add1=b[i].add2=b[i].step=0;}}void updateA(ll l,ll r,ll add,ll i){ll mid,add1,add2;if(b[i].l==l && b[i].r==r){add1=add;add2=add+b[i].r-b[i].l;b[i].add1+=add1;b[i].add2+=add2;b[i].step+=1;b[i].sum+=(b[i].r-b[i].l+1)*(add2+add1)/2;//b[i].flag=-1;这里注意不能使得b[i].flag=-1,因为对于所有数相同的区间，它的add1,add2可以不为0 return;}pushdown(i);mid=(b[i].l+b[i].r)/2;if(r<=mid)updateA(l,r,add,i*2);else if(l>mid)updateA(l,r,add,i*2+1);else{updateA(l,mid,add,i*2);updateA(mid+1,r,add+mid+1-l,i*2+1);}b[i].sum=b[i*2].sum+b[i*2+1].sum;}void updateB(ll l,ll r,ll add,ll i){ll mid,add1,add2;if(b[i].l==l && b[i].r==r){add1=add+b[i].r-b[i].l;add2=add;b[i].add1+=add1;b[i].add2+=add2;b[i].step-=1;b[i].sum+=(b[i].r-b[i].l+1)*(add2+add1)/2;return;}pushdown(i);mid=(b[i].l+b[i].r)/2;if(r<=mid)updateB(l,r,add,i*2);else if(l>mid)updateB(l,r,add,i*2+1);else{updateB(l,mid,add+r-mid,i*2);updateB(mid+1,r,add,i*2+1);}b[i].sum=b[i*2].sum+b[i*2+1].sum;}void updateC(ll l,ll r,ll num,ll i){ll mid;if(b[i].l==l && b[i].r==r){b[i].add1=b[i].add2=b[i].step=0;b[i].sum=(b[i].r-b[i].l+1)*num;b[i].flag=0;b[i].v=num;return;}pushdown(i);mid=(b[i].l+b[i].r)/2;if(r<=mid)updateC(l,r,num,i*2);else if(l>mid)updateC(l,r,num,i*2+1);else{updateC(l,mid,num,i*2);updateC(mid+1,r,num,i*2+1);}b[i].sum=b[i*2].sum+b[i*2+1].sum;}void question(ll l,ll r,ll i){ll mid;if(b[i].l==l && b[i].r==r){sum+=b[i].sum;return;}pushdown(i);mid=(b[i].l+b[i].r)/2;if(r<=mid) question(l,r,i*2);else if(l>mid) question(l,r,i*2+1);else{question(l,mid,i*2);question(mid+1,r,i*2+1);}}int main(){ll n,m,i,j;while(scanf("%lld",&n)!=EOF){build(1,250000,1);for(i=1;i<=n;i++){char s[10];ll c,d,e;scanf("%s",s);if(s[0]=='A'){scanf("%lld%lld",&c,&d);updateA(c,d,1,1);}else if(s[0]=='B'){scanf("%lld%lld",&c,&d);updateB(c,d,1,1);}else if(s[0]=='C'){scanf("%lld%lld%lld",&c,&d,&e);updateC(c,d,e,1);}else if(s[0]=='S'){scanf("%lld%lld",&c,&d);sum=0;question(c,d,1);printf("%lld\n",sum);}}}return 0;}/*10B 1 4A 1 4S 2 2B 1 3C 1 4 1B 1 4C 3 3 3A 2 3S 2 3B 3 3*/
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勇士面前无险路。
