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Case 1: The total value of the hook is 24.
//类似延迟标记一样的思路
//#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100011;
const int inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,n) for(int i=0;i<(n);i++)
template<class T>inline T read(T&x)
{
char c;
while((c=getchar())<=32);
bool ok=false;
if(c=='-')ok=true,c=getchar();
for(x=0; c>32; c=getchar())
x=x*10+c-'0';
if(ok)x=-x;
return x;
}
template<class T> inline void read_(T&x,T&y)
{
read(x);
read(y);
}
template<class T> inline void write(T x)
{
if(x<0)putchar('-'),x=-x;
if(x<10)putchar(x+'0');
else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
write(x);
putchar('\n');
}
//-------IO template------
typedef long long LL;
struct node
{
int sum;
int val;
} p[maxn<<3];
void build(int rt,int L,int R)
{
p[rt].val=0;
if(L==R)
{
p[rt].sum=1;
p[rt].val=1;
return;
}
build(lson);
build(rson);
p[rt].sum=p[rt<<1].sum+p[rt<<1|1].sum;
}
void update(int rt,int L,int R,int x,int y,int z)
{
// printf("%d %d %d %d %d %d\n",rt,L,R,x,y,z);
if(x==L&&y==R)
{
p[rt].sum=(R-L+1)*z;
p[rt].val=z;
return ;
}
if(p[rt].val>0)//向下更新时，判断条件写好，
{
p[rt<<1].sum=(M-L+1)*p[rt].val;
p[rt<<1|1].sum=(R-M)*p[rt].val;//右儿子 区间是R-M即R-（M+1）+1
p[rt<<1].val=p[rt<<1|1].val=p[rt].val;
p[rt].val=0;
}
if(y<=M)
update(lson,x,y,z);
else if(x>M)
update(rson,x,y,z);
else
{
update(lson,x,M,z);
update(rson,M+1,y,z);
}
p[rt].sum=p[rt<<1].sum+p[rt<<1|1].sum;
}
int main()
{
//#ifndef ONLINE_JUDGE
// freopen("in.txt","r",stdin);
//#endif // ONLINE_JUDGE
int T;
int n,m,i,j,k,t;
read(T);
int cas=1;
while(T--)
{
read_(n,m);
build(1,1,n);
int x,y,z;
while(m--)
{
read_(x,y);
read(z);
update(1,1,n,x,y,z);
}
printf("Case %d: The total value of the hook is %d.\n",cas++,p[1].sum);
}
return 0;
}
E - Just a Hook HDU 1698 （线段树+类似延迟标记）
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原文地址：http://blog.csdn.net/u013167299/article/details/44874513
