HDU 1698 Just a Hook (线段树)

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11425    Accepted Submission(s): 5650

Problem 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.

 

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.

 

Output

For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.

 

Sample Input

1 10 2 1 5 2 5 9 3

 

Sample Output

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

 

Source

2008 “Sunline Cup” National Invitational Contest

 

Recommend

wangye

题意：dota中的英雄 Pudge 有一hook 这hook 可由三种材料构成 分别如下
For each cupreous stick, the value is 1.
For each silver stick, the value is 2.
For each golden stick, the value is 3.
现输入 x y z 表示x到y的hook 的值变为z 。z如上如述
求最后，hook的值

思路：线段树 普通的线段树会超时，得成段更新 用到了延迟覆盖

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=100010;

#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)

struct Tree{
    int l,r;
    int val,sum;
    int cover;
}tree[N<<2];

void PushUp(int rt){
    tree[rt].sum=tree[L(rt)].sum+tree[R(rt)].sum;
}

void build(int L,int R,int rt){
    tree[rt].l=L;
    tree[rt].r=R;
    tree[rt].cover=1;
    tree[rt].val=1;
    if(tree[rt].l==tree[rt].r){
        tree[rt].sum=1;
        return ;
    }
    int mid=(L+R)>>1;
    build(L,mid,L(rt));
    build(mid+1,R,R(rt));
    PushUp(rt);
}

void PushDown(int rt){
    tree[L(rt)].val=tree[rt].val;
    tree[L(rt)].sum=(tree[L(rt)].r-tree[L(rt)].l+1)*tree[rt].val;
    tree[L(rt)].cover=1;
    tree[R(rt)].val=tree[rt].val;
    tree[R(rt)].sum=(tree[R(rt)].r-tree[R(rt)].l+1)*tree[rt].val;
    tree[R(rt)].cover=1;
    tree[rt].cover=0;
}

void update(int val,int L,int R,int rt){
    if(tree[rt].l==L && tree[rt].r==R){
        tree[rt].cover=1;
        tree[rt].val=val;
        tree[rt].sum=(tree[rt].r-tree[rt].l+1)*val;
        return ;
    }
    if(tree[rt].cover)
        PushDown(rt);
    if(tree[rt].val==val && tree[rt].cover) //当这一整段已为val,且覆盖整段，则继续向下已没有意义
        return ;
    int mid=(tree[rt].l+tree[rt].r)>>1;
    if(R<=mid)
        update(val,L,R,L(rt));
    else if(L>=mid+1)
        update(val,L,R,R(rt));
    else{
        update(val,L,mid,L(rt));
        update(val,mid+1,R,R(rt));
    }
    PushUp(rt);
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,n,q;
    int cases=0;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&q);
        build(1,n,1);
        int a,b,z;
        while(q--){
            scanf("%d%d%d",&a,&b,&z);
            update(z,a,b,1);
        }
        printf("Case %d: The total value of the hook is %d.\n",++cases,tree[1].sum);
    }
    return 0;
}