点击打开链接题目链接

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.

金银铜价值分别为3 2 1

屠夫的钩子刚開始是铜做的

给出n长度的钩子 q次操作将a b直接的钩子替换成价值为c的金属

问最后价值

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=111111;
int sum[MAXN<<2];
int lazy[MAXN<<2];
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
    lazy[rt]=0;
    if(l==r)
    {
        sum[rt]=1;
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    pushup(rt);
}
void pushdown(int rt,int m)
{
    if(lazy[rt])
    {
        lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];
        sum[rt<<1]=(m-(m>>1))*lazy[rt];
        sum[rt<<1|1]=(m>>1)*lazy[rt];
        lazy[rt]=0;
    }
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(l>=L&&R>=r)
    {
        lazy[rt]=c;
        sum[rt]=(r-l+1)*c;
        return ;
    }
    pushdown(rt,r-l+1);
    int mid=(l+r)>>1;
    if(L<=mid)
        update(L,R,c,l,mid,rt<<1);
    if(R>mid)
        update(L,R,c,mid+1,r,rt<<1|1);
    pushup(rt);
}
int main()
{
    int t,n,q,a,b,c;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        scanf("%d %d",&n,&q);
        build(1,n,1);
        while(q--)
        {
            scanf("%d %d %d",&a,&b,&c);
            update(a,b,c,1,n,1);
        }
        printf("Case %d: The total value of the hook is %d.\n",cas,sum[1]);
    }
    return 0;
}