(简单) HDU 1698 Just a Hook , 线段树+区间更新。

Description：　　

　　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.

　　

　　题目大意就是给你一排棒子，初始都是铁的，然后改变其中一部分为别的，最后问总价值和。

　　典型的线段树区间更新问题。

 

代码如下：

#include<iostream>
#include<cstdio>

#define lson L,M,po*2
#define rson M+1,R,po*2+1

using namespace std;

int COL[100005*4];
int BIT[100005*4];

void pushUP(int po)
{
    BIT[po]=BIT[po*2]+BIT[po*2+1];
}

void pushDown(int po,int len)
{
    if(COL[po])
    {
        BIT[po*2]=(len-(len/2))*COL[po];
        BIT[po*2+1]=(len/2)*COL[po];
        COL[po*2]=COL[po*2+1]=COL[po];
        COL[po]=0;
    }
}

void build_tree(int L,int R,int po)
{
    BIT[po]=(R-L+1);
    COL[po]=0;

    if(L==R)
        return;

    int M=(L+R)/2;

    build_tree(lson);
    build_tree(rson);
}

void update(int ul,int ur,int type,int L,int R,int po)
{
    if(ul<=L&&ur>=R)
    {
        BIT[po]=type*(R-L+1);
        COL[po]=type;

        return;
    }

    pushDown(po,(R-L+1));

    int M=(L+R)/2;

    if(ul<=M)
        update(ul,ur,type,lson);
    if(ur>M)
        update(ul,ur,type,rson);

    pushUP(po);
}

int main()
{
    int T;
    int N,M;
    int a,b,t;
    cin>>T;

    for(int ii=1;ii<=T;++ii)
    {
        scanf("%d",&N);
        build_tree(1,N,1);

        scanf("%d",&M);

        for(int i=0;i<M;++i)
        {
            scanf("%d %d %d",&a,&b,&t);
            update(a,b,t,1,N,1);
        }

        printf("Case %d: The total value of the hook is %d.\n",ii,BIT[1]);
    }

    return 0;
}