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Just a Hook

In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.



Now Pudge wants to do some operations on the hook.

Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:

For each cupreous stick, the value is 1.
For each silver stick, the value is 2.
For each golden stick, the value is 3.

Pudge wants to know the total value of the hook after performing the operations.
You may consider the original hook is made up of cupreous sticks.

1
10
2
1 5 2
5 9 3

Case 1: The total value of the hook is 24.

题意很简单，求总区间的和。每次更新时不要更新到底层就行，更新完成时顺便求和就行。最后f[1].s即是ans.

#include<cstdio>
#include<cmath>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll __int64
#define N 200005
struct node
{
    int l,r;
    int s,v,f;  //区间和、该区间的单个点的值、f为不零则该区间有待更新值
}f[N*3];
void creat(int t,int l,int r)
{
    f[t].l=l;
    f[t].r=r;
    f[t].v=1;
    f[t].f=0;
    if(l==r)
    {
        f[t].s=1;
        return ;
    }
    int tmp=t<<1,mid=(l+r)>>1;
    creat(tmp,l,mid);
    creat(tmp|1,mid+1,r);
    f[t].s=f[tmp].s+f[tmp|1].s;
}
void update(int t,int l,int r,int v)
{
    int tmp=t<<1,mid=(f[t].l+f[t].r)>>1;
    if(f[t].l==l&&f[t].r==r)
    {
        f[t].v=f[t].f=v;      
        f[t].s=(r-l+1)*v;
        return ;
    }
    if(f[t].f)      //向下更新
    {
        f[tmp].f=f[tmp|1].f=f[t].f;
        f[tmp].v=f[tmp|1].v=f[t].f;
        f[tmp].s=(f[tmp].r-f[tmp].l+1)*f[t].f;
        f[tmp|1].s=(f[tmp|1].r-f[tmp|1].l+1)*f[t].f;
        f[t].f=0;
    }
    if(r<=mid)
        update(tmp,l,r,v);
    else if(l>mid)
        update(tmp|1,l,r,v);
    else
    {
        update(tmp,l,mid,v);
        update(tmp|1,mid+1,r,v);
    }
    f[t].s=f[tmp].s+f[tmp|1].s;      //向上求和
    f[t].f=0;
}
int main()
{
    int T,i,n,q,cnt=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&q);
        creat(1,1,n);
        while(q--)
        {
            int l,r,v;
            scanf("%d%d%d",&l,&r,&v);
            update(1,l,r,v);
        }
        printf("Case %d: The total value of the hook is %d.\n",cnt++,f[1].s);
    }
    return 0;
}

hdu 1698 Just a Hook (成段更新+总区间求和)

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原文地址：http://blog.csdn.net/u011721440/article/details/41629659