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1、
这道题目就是错在懒惰标记没有处理好,而且没有注意long long
懒惰标记实际上就是让子节点暂时处于不更新状态,用到的时候再更新,如本题中的visit[]就是懒惰标记,例如总长度是1-10,我们现在要想更新1-6,(将1-6的值都加3)那么update()会先找1-10,发现不合适,再找他的左右孩子,发现1<5,说明1-6的区间在1-10的左孩子中,同时6>5,1-6也在1-10的右孩子中,这样依次去找1-6在的区间。但是找到1-5的时候,我们发现整个1-5都在1-6中间,也就是说这一段都要更新,那么我们将1-5的sum值更新了,同时用visit[rt]+=3记录下来1-5中的数字现在每个都 要加的数字,但是1-5下边还有1-3,4-5,3-3,4-4,5-5,这些我们就可以不用更新,因为这些我们暂时还用不到,假如现在又要将1-5区间的值都加5,那么visit[rt]+=5,此时就是8了,但是还是不用更新他的子节点,假如我们现在要用到1-3区间了,我们就可以一次性给1-3区间加上8,而不用先加3,再加5,这样懒惰标记就使得每次的递归都少了好多,
2、题目:
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 52972 | Accepted: 15828 | |
Case Time Limit: 2000MS |
Description
You have N integers, A1,A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1,A2, ... , AN. -1000000000 ≤Ai ≤ 1000000000. Each of the next Q lines represents an operation. "C a b c" means adding c to each of Aa,Aa+1, ... ,Ab. -10000 ≤ c ≤ 10000. "Q a b" means querying the sum of Aa,Aa+1, ... ,Ab.Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4
Sample Output
455915
Hint
Source
#include#include #include using namespace std;#define N 100005#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ll long longll sum[N*4];int visit[N*4];void pushUp(int rt){ sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt){ visit[rt]=0; if(l==r) { scanf("%lld",&sum[rt]); return ; } int m=(l+r)>>1; build(lson); build(rson); pushUp(rt);}void pushDown(int rt,int d){ if(visit[rt]!=0) { visit[rt<<1]+=visit[rt]; visit[rt<<1|1]+=visit[rt]; sum[rt<<1|1]+=(ll)(d>>1)*visit[rt];//注意后边乘可能超整形,强制类型转换(ll),即AC sum[rt<<1]+=(ll)(d-(d>>1))*visit[rt]; visit[rt]=0; }}void update(int L,int R,int c,int l,int r,int rt){ if(L<=l && R>=r) { visit[rt]+=c; sum[rt]+=(r-l+1)*c; return ; } pushDown(rt,r-l+1); int m=(l+r)>>1; if(L<=m) update(L,R,c,lson); if(R>m) update(L,R,c,rson); pushUp(rt);}ll query(int L,int R,int l,int r,int rt){ if(L<=l && R>=r) { return sum[rt]; } pushDown(rt,r-l+1); int m=(l+r)>>1; ll ret=0; if(L<=m) ret+=query(L,R,lson); if(R>m) ret+=query(L,R,rson); return ret;}int main(){ int n,q,a,b,c; char s[3]; scanf("%d%d",&n,&q); build(1,n,1); while(q--) { scanf("%s",s); if(s[0]=='C') { scanf("%d%d%d",&a,&b,&c); update(a,b,c,1,n,1); } else if(s[0]=='Q') { scanf("%d%d",&a,&b); printf("%lld\n",query(a,b,1,n,1)); } } return 0;}/*10 1001 2 3 4 5 6 7 8 9 10C 1 6 3Q 1 1Q 1 2Q 4 4Q 1 10Q 1 5C 3 6 3Q 1 5*/
4、wrong answer 代码:
#include#include #include using namespace std;#define N 100005#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ll long longll sum[N*4];int visit[N*4];void pushUp(int rt){ sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt){ visit[rt]=0; if(l==r) { scanf("%lld",&sum[rt]); return ; } int m=(l+r)>>1; build(lson); build(rson); pushUp(rt);}void pushDown(int L,int R,int c,int l,int r,int rt){ int m=r-l+1; if(visit[rt]!=0) { //printf("%d %d %d %d %d %d\n",l,r,sum[rt<<1|1],sum[rt<<1],m,c); sum[rt<<1|1]+=(m>>1)*c; sum[rt<<1]+=(m-(m>>1))*c; // printf("&&&&&&%d %d\n",sum[rt<<1|1],sum[rt<<1]); visit[rt]=0; }}void update(int L,int R,int c,int l,int r,int rt){ //printf("*%d %d %d \n",l,r,rt); if(L<=l && R>=r) { sum[rt]+=(r-l+1)*c; visit[rt]=1; pushDown(L,R,c,l,r,rt); return ; } pushDown(L,R,c,l,r,rt); int m=(l+r)>>1; if(L<=m) update(L,R,c,lson); if(R>m) update(L,R,c,rson); pushUp(rt);}ll query(int L,int R,int l,int r,int rt){ if(L<=l && R>=r) { return sum[rt]; } int m=(l+r)>>1; ll ret=0; if(L<=m) ret+=query(L,R,lson); if(R>m) ret+=query(L,R,rson); return ret;}int main(){ int n,q,a,b,c; char s[3]; scanf("%d%d",&n,&q); build(1,n,1); while(q--) { scanf("%s",s); if(s[0]=='C') { scanf("%d%d%d",&a,&b,&c); update(a,b,c,1,n,1); } else if(s[0]=='Q') { scanf("%d%d",&a,&b); printf("%lld\n",query(a,b,1,n,1)); } } return 0;}/*10 1001 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 1 5C 3 6 3Q 1 5*/
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