题目链接:http://poj.org/problem?id=3468
题目思路:区间修改,区间查询,线段树模板题
代码:
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>
using namespace std;
#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)
const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const double DINF = 0xffffffffffff;
const int mod = 1e9+7;
const int N = 1e6+5;
ll a[N],n; //输入数据,构造线段树数组
struct node{
int lft,rht;
ll sum; //区间和
ll lazy; //延迟标记,减小时间复杂度
}segTree[N*4]; //需要开4倍
void pushUp(int id){ //区间合并,上放
segTree[id].sum = segTree[id*2].sum+segTree[id*2+1].sum;
}
void pushDown(int id){
if(segTree[id].lazy){ //区间修改过,需要下放
//在原来的值上加上val
segTree[id*2].sum += (segTree[id*2].rht-segTree[id*2].lft+1)*segTree[id].lazy;
segTree[id*2+1].sum += (segTree[id*2+1].rht-segTree[id*2+1].lft+1)*segTree[id].lazy;
segTree[id*2].lazy += segTree[id].lazy;
segTree[id*2+1].lazy += segTree[id].lazy;
segTree[id].lazy = 0;
}
}
void build(int id,int l,int r){
segTree[id].lft = l, segTree[id].rht = r;
segTree[id].lazy = 0, segTree[id].sum = 0; //开始一定要清0
if(l == r){ //到达叶子节点,不继续建树
segTree[id].sum = a[l];
}
else{ //否则继续建树
int mid = (l+r)>>1;
build(id*2,l,mid);
build(id*2+1,mid+1,r);
pushUp(id);
}
}
void upDate(int id,int l,int r,int val){ //更新l~r区间,加val,或减val(就传-val),或改成val
if(l<=segTree[id].lft&&r>=segTree[id].rht){
/*1.把原来的值加上val,因为该区间有segTree[id].rht-segTree[index].lft+1
个数,所以区间和 以及 最值为:*/
segTree[id].sum += (segTree[id].rht-segTree[id].lft+1)*val;
segTree[id].lazy += val; //延迟标记
}
else{
pushDown(id); //区间下放
int mid = (segTree[id].lft+segTree[id].rht)>>1;
if(r <= mid)
upDate(id*2,l,r,val);
else if(l>=mid+1)
upDate(id*2+1,l,r,val);
else{
upDate(id*2,l,r,val);
upDate(id*2+1,l,r,val);
}
pushUp(id);
}
}
ll query(int id,int l,int r){ //查询l~r的值
if(l<=segTree[id].lft&&r>=segTree[id].rht){ //该区间包含在查询区间内,可直接返回
return segTree[id].sum;
}
pushDown(id); //区间下放
int mid = (segTree[id].lft+segTree[id].rht)>>1;
ll ans = 0;
if(r<=mid){ //只用管左子树
ans += query(id*2,l,r);
}
else if(l>=mid+1){ //只用管右子树
ans+=query(id*2+1,l,r);
}
else{
ans += query(id*2,l,r)+query(id*2+1,l,r);
}
return ans;
//return maxx;
//return minn;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
std::ios::sync_with_stdio(false);
ll t,l,r,x,m;
char op;
while(~scanf("%lld%lld",&n,&m)){
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
build(1,1,n);
while(m--){
scanf(" %c",&op);
if(op=='Q')
scanf("%lld%lld",&l,&r),printf("%lld\n",query(1,l,r));
else
scanf("%lld%lld%lld",&l,&r,&x),upDate(1,l,r,x);
}
}
return 0;
}
树状数组写法,树状数组区间写法详解可看:https://blog.csdn.net/baodream/article/details/80207879
本题AC代码:
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>
using namespace std;
#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)
const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const double DINF = 0xffffffffffff;
const int mod = 1e9+7;
const int N = 1e6+5;
ll a[N],tree[N],tree2[N]; //因为tree2是乘积,有可能超int
ll n;
int lowbit(int x){return x&(-x);}
void updata(ll c[],int x,ll val){ //单点更新
while(x<=n){
c[x]+=val;
x+=lowbit(x);
}
}
ll query(const ll c[],ll x){ //1~x前缀和查询
ll res = 0;
while(x>0){
res+=c[x];
x-=lowbit(x);
}
return res;
}
void regionUpdata(ll x,ll y,ll val){ //区间更新,实现a[x]~a[y]+val
updata(tree,x,val);
updata(tree,y+1,-val);
updata(tree2,x,x*val);
updata(tree2,y+1,-(y+1)*val);
}
ll regionQuery(ll x,ll y){
//求1~y的和
ll ans1 = (y+1)*query(tree,y) - query(tree2,y);
//求1~x-1的和
ll ans2 = x*query(tree,x-1) - query(tree2,x-1);
return ans1-ans2;
}
void init(){
a[0]=0;
for(ll i=1;i<=n;i++){
updata(tree,i,a[i]-a[i-1]); //用差分数组初始化树状数组
updata(tree2,i,i*(a[i]-a[i-1]));
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
std::ios::sync_with_stdio(false);
ll q,x,y,val;
char typ;
scanf("%lld",&n);scanf("%lld",&q);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
init();
while(q--){
scanf(" %c",&typ);
if(typ=='C'){
scanf("%lld%lld%lld",&x,&y,&val);
regionUpdata(x,y,val);
}
else{
scanf("%lld%lld",&x,&y);
printf("%lld\n",regionQuery(x,y));
}
}
return 0;
}