HDU1698 - Just a Hook - 线段树之区间更新

1.题目描述：

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31072    Accepted Submission(s): 15307

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

 

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2.题意概述：

x a b就是把a ~ b区间的值都改成x

3.解题思路：

也是一个线段树的裸题，可以用lazy降低复杂度，但是好像这题不明显。

4.AC代码：

#include 
#define INF 0x3f3f3f3f
#define maxn 100010
#define lson root << 1
#define rson root << 1 | 1
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
struct tree
{
    int l, r, w, lazy;
} t[maxn << 2];
void pushup(int root)
{
    t[root].w = t[lson].w + t[rson].w;
}
void pushdown(int root)
{
    if (t[root].lazy)
    {
        t[lson].lazy = t[rson].lazy = t[root].lazy;
        t[lson].w = (t[lson].r - t[lson].l + 1) * t[root].lazy;
        t[rson].w = (t[rson].r - t[rson].l + 1) * t[root].lazy;
        t[root].lazy = 0;
    }
}
void build(int l, int r, int root)
{
    t[root].l = l;
    t[root].r = r;
    t[root].w = 0;
    t[root].lazy = 0;
    if (l == r)
    {
        t[root].w = 1;
        return;
    }
    int mid = l + r >> 1;
    build(l, mid, lson);
    build(mid + 1, r, rson);
    pushup(root);
}
void update(int z, int l, int r, int root)
{
    if (l <= t[root].l && t[root].r <= r)
    {
        t[root].lazy = z;
        t[root].w = z * (t[root].r - t[root].l + 1);
        return;
    }
    pushdown(root);
    int mid = t[root].l + t[root].r >> 1;
    if (l <= mid)
        update(z, l, r, lson);
    if (r > mid)
        update(z, l, r, rson);
    pushup(root);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    long _begin_time = clock();
#endif
    int T, n, q, kase = 0;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &q);
        build(1, n, 1);
        while (q--)
        {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            update(z, x, y, 1);
        }
        printf("Case %d: The total value of the hook is %d.\n", ++kase, t[1].w);
    }
#ifndef ONLINE_JUDGE
    long _end_time = clock();
    printf("time = %ld ms.", _end_time - _begin_time);
#endif
    return 0;
}