HDOJ-1698-线段树成段更新

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22882    Accepted Submission(s): 11459

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.

 

题目大意：初始给定n个数，每个原始值为1，然后对这些数进行操作，把[区间[  X ,Y  ]的值更新为Z,问这些数最后的和为多少，可以用到线段树成段更新，设置一个 col 数组，每次更新此节点的时候，将col值设置为更新的值，而不继续对子区间进行更新，做到缓冲的效果，而要访问子区间的时候再将保存的值传递给子区间。

#include<iostream>
#include<algorithm>
#include<cstdio>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define MAXN 333333
using namespace std;
int sum[MAXN<<2];
int col[MAXN<<2];
void PushUp(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void PushDown(int rt,int m)
{
    if(col[rt]){
        col[rt<<1]=col[rt<<1|1]=col[rt];
        sum[rt<<1]=(m-(m>>1))*col[rt];
        sum[rt<<1|1]=(m>>1)*col[rt];
        col[rt]=0;
    }
}
void build(int l,int r,int rt){
    col[rt]=0;
    sum[rt]=1;
    if(l==r)return;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    PushUp(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R){
        col[rt]=c;
        sum[rt]=c*(r-l+1);
        return ;
    }
    PushDown(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)update(L,R,c,lson);
    if(R>m)update(L,R,c,rson);
    PushUp(rt);
}
int main()
{
    int T,N,Q,i,j,c,cnt=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        build(1,N,1);
        scanf("%d",&Q);
        while(Q--)
        {
            scanf("%d%d%d",&i,&j,&c);
            update(i,j,c,1,N,1);
        }
        printf("Case %d: The total value of the hook is %d.\n",cnt++,sum[1]);
    }
}

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