HDU – 1698 Just a Hook（[线段树] 区间更新+区间和查询）

#include<algorithm>
#include <iostream>
#include   <fstream>
#include  <cstdlib>
#include  <cstring>
#include  <cassert>
#include   <cstdio>
#include   <vector>
#include   <string>
#include    <cmath>
#include    <queue>
#include    <stack>
#include      <set>
#include      <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000007;
const int INF = 0x3f3f3f3f;
ll gcd(ll a, ll b) {
	return b ? gcd(b, a%b) : a;
}
template<class T>inline void gmax(T &A, T B) {
	(A<B)&&(A=B);//if(B>A)A=B;
}
template<class T>inline void gmin(T &A, T B) {
	(A>B)&&(A=B);//if(B<A)A=B;
}
template <class T>
inline bool scan_d(T &ret) {
	char c;
	int sgn;
	if(c=getchar(),c==EOF) return 0; //EOF
	while(c!='-'&&(c<'0'||c>'9')) c=getchar();
	sgn=(c=='-')?-1:1;
	ret=(c=='-')?0:(c-'0');
	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
	ret*=sgn;
	return 1;
}
inline void outt(int x) {
	if(x>9) outt(x/10);
	putchar(x%10+'0');
}
const int maxn = 1e6+10;
struct SegmentTree {
	int l,r;
	long long sum, add;
#define l(x) tree[x].l
#define r(x) tree[x].r
#define sum(x) tree[x].sum
#define add(x) tree[x].add
} tree[1000010*4];
int a[1000010],n,m;
void build(int p,int l,int r) {
	l(p)=l,r(p)=r;
	add(p)=0;//注意初始化位置 
	if(l==r) {
		sum(p)=a[l];
		return ;
	}
	int mid=(l+r)/2;//只有这里的一半是直接用传入的l，r，但实际上这里的l，r就是下面的p.l,p.r
	build(p*2,l,mid);
	build(p*2+1,mid+1,r);
	sum(p)=sum(p*2)+sum(p*2+1);
}
//延迟标记代表：该节点已经被修改，但子节点尚未被更新
void spread(int p) {
	if(add(p)) {
		sum(p*2) = add(p)*(r(p*2)-l(p*2)+1); //更新左子节点信息
		sum(p*2+1) = add(p)*(r(p*2+1)-l(p*2+1)+1);//更新右子节点信息
		add(p*2) = add(p); //给左子节点打延迟标记
		add(p*2+1) = add(p);
		add(p) = 0;//清楚p节点的标记
	}
}
void change(int p,int l,int r,int d) {
	if(l<=l(p) && r>=r(p)) { //完成覆盖
		sum(p)=(long long)d*(r(p)-l(p)+1);//更新节点信息
		add(p)=d;//给节点打延迟标记
		return ;
	}
	spread(p); //下传延迟标记
	int mid=(l(p)+r(p))/2; //这里的中间是当前节点的中间（或者说是偏左的中间）位置
	if(l<=mid)change(p*2,l,r,d);
	if(r>mid)change(p*2+1,l,r,d);//注意这里不是非此即彼的关系
	sum(p)=sum(p*2)+sum(p*2+1);
}
long long ask(int p,int l,int r) {
	if(l<=l(p) && r>=r(p))return sum(p);
	spread(p);
	int mid=(l(p)+r(p))/2;
	long long val=0;
	if(l <= mid) val+=ask(p*2,l,r);//注意！这里对中间元素进行比较的是l不是tree[p].l
	if(r > mid) val+=ask(p*2+1,l,r);
	return val;
}
char cmd[10];
int main() {
	int t;
	cin>>t;
	rep(l,1,t) {
		int n,m;
		scan_d(n),scan_d(m);
		rep(i,1,n)a[i]=1;
		build(1,1,n);
		while(m--) {
			int l,r,d;
			scan_d(l);
			scan_d(r);
			scan_d(d);
			change(1,l,r,d);
		}
		printf("Case %d: The total value of the hook is %d.\n",l,ask(1,1,n));
	}
	return 0;
}