POJ 3468 A Simple Problem with Integers（線段樹或樹狀數組）

題目鏈接：http://poj.org/problem?id=3468

題目思路：區間修改，區間查詢，線段樹模板題

代碼：

#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;  //輸入數據,構造線段樹數組

struct node{
    int lft,rht;
    ll sum;    //區間和
    ll lazy;   //延遲標記,減小時間複雜度
}segTree[N*4];  //需要開4倍

void pushUp(int id){    //區間合併,上放
    segTree[id].sum = segTree[id*2].sum+segTree[id*2+1].sum;
}

void pushDown(int id){
    if(segTree[id].lazy){   //區間修改過,需要下放
        //在原來的值上加上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;   //開始一定要清0
    if(l == r){            //到達葉子節點,不繼續建樹
        segTree[id].sum = a[l];
    }
    else{                  //否則繼續建樹
        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區間,加val,或減val(就傳-val),或改成val
    if(l<=segTree[id].lft&&r>=segTree[id].rht){
        /*1.把原來的值加上val,因爲該區間有segTree[id].rht-segTree[index].lft+1
        個數，所以區間和 以及 最值爲：*/
        segTree[id].sum += (segTree[id].rht-segTree[id].lft+1)*val;
        segTree[id].lazy += val;  //延遲標記
    }
    else{
        pushDown(id);   //區間下放
        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){  //查詢l~r的值
    if(l<=segTree[id].lft&&r>=segTree[id].rht){  //該區間包含在查詢區間內，可直接返回
        return segTree[id].sum;
    }
    pushDown(id);  //區間下放
    int mid = (segTree[id].lft+segTree[id].rht)>>1;
    ll ans = 0;
    if(r<=mid){  //只用管左子樹
        ans += query(id*2,l,r);
    }
    else if(l>=mid+1){  //只用管右子樹
        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;
}

樹狀數組寫法，樹狀數組區間寫法詳解可看：https://blog.csdn.net/baodream/article/details/80207879

本題AC代碼：

#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]; //因爲tree2是乘積，有可能超int
ll n;
int lowbit(int x){return x&(-x);}
void updata(ll c[],int x,ll val){  //單點更新
    while(x<=n){
        c[x]+=val;
        x+=lowbit(x);
    }
}
ll query(const ll c[],ll x){   //1~x前綴和查詢
    ll res = 0;
    while(x>0){
        res+=c[x];
        x-=lowbit(x);
    }
    return res;
}
void regionUpdata(ll x,ll y,ll val){ //區間更新，實現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]);   //用差分數組初始化樹狀數組
        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;
}