POJ 3468 A Simple Problem with Integers 线段树 区间更新

原题： http://poj.org/problem?id=3468

题目：

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 78207 Accepted: 24097
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 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output

4
55
9
15
Hint

The sums may exceed the range of 32-bit integers.

思路：

直接套模版。
注意数据会int溢出。

代码：

#include <iostream>#include"cstdio"#include"string.h"using namespace std;typedef long long int lint;const int N= 100005;struct node{    int left;    int right;    lint add;    lint sum;} tree[N*4];int n,m;lint a[N];void init(){    memset(tree,0,sizeof(tree));    memset(a,0,sizeof(a));}//将下面更新的值返回上一层void pushup(int id){    tree[id].sum=tree[id*2].sum+tree[id*2+1].sum;}//将标记区间向下移动void pushdown(int id){    tree[id*2].add=tree[id*2].add+tree[id].add;    tree[id*2].sum=tree[id*2].sum+(tree[id*2].right-tree[id*2].left+1)*tree[id].add;    tree[id*2+1].add=tree[id*2+1].add+tree[id].add;    tree[id*2+1].sum=tree[id*2+1].sum+(tree[id*2+1].right-tree[id*2+1].left+1)*tree[id].add;    tree[id].add=0;}//建树的时候每次建完下一层要向上层赋值void build(int id,int l,int r){    tree[id].left=l;    tree[id].right=r;    tree[id].add=0;    tree[id].sum=0;    if(l==r)    {        tree[id].sum=a[l];        return;    }    int mid=(l+r)/2;    build(id*2,l,mid);    build(id*2+1,mid+1,r);    pushup(id);}void add(int id,int l,int r,lint val){    //如果当前区间完全被包含，只更新当前区间的总和    //并对该区间做增加标记    if(tree[id].left>=l&&tree[id].right<=r)    {        tree[id].add=tree[id].add+val;        tree[id].sum=tree[id].sum+(tree[id].right-tree[id].left+1)*val;        return ;    }    //如果当前区间没有被包含    if(tree[id].right<l||tree[id].left>r)   return ;    //如果当前区间部分被包含，标记下移    if(tree[id].add)    pushdown(id);    //更新左右子区间    add(id*2,l,r,val);    add(id*2+1,l,r,val);    //更新完再把结果返回上层    pushup(id);}lint ans;void query(int id,int l,int r){    //查询区间在该区间外    if(tree[id].left>r||tree[id].right<l)   return ;    //该区间完全被包含    if(tree[id].left>=l&&tree[id].right<=r)    {        ans=ans+tree[id].sum;        return ;    }    //查询部分该区间，标记下移    if(tree[id].add)    pushdown(id);    //左右查找    int mid=(tree[id].left+tree[id].right)/2;    if(l<=mid)  query(id*2,l,r);    if(r>mid)   query(id*2+1,l,r);}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d %d",&n,&m)!=EOF)    {        init();        char s[10];        for(int i=1; i<=n; i++)            scanf("%I64d",&a[i]);        build(1,1,n);        for(int i=1; i<=m; i++)        {            int a,b,c;            scanf("%s",s);            if(s[0]=='Q')            {                ans=0;                scanf("%d %d",&a,&b);                query(1,a,b);                printf("%I64d\n",ans);            }            if(s[0]=='C')            {                scanf("%d %d %d",&a,&b,&c);                add(1,a,b,c);            }        }    }    //fclose(stdin);    return 0;}