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hdu1698 Just a Hook（線段樹）

編輯：關於C語言

hdu1698 Just a Hook（線段樹）

Just a Hook

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 24473 Accepted Submission(s): 12193

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

Sample Output

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

Source 2008 “Sunline Cup” National Invitational Contest
題意：一段線段由n條小線段組成，每次操作把一個區間的小線段變成金銀銅之一（金的價值為3，銀為2，銅為1），最初可當做全為銅；最後求這條線段的總價值。分析：基礎的區間更新，我也是剛學，按照我的理解區間更新就是，樹最開始有個狀態，然後每次更新就相當於把一個節點為根節點的樹全部變成同一種狀態；也就是每個結點有個數，表示以該結點為根結點的樹的所有結點都是這個數，或者表示以該結點為根的樹有不同數，那麼就把他變為-1（或者其他，你開心就好），然後再往下處理。不知道能不能看懂，真的不太好描述- -。

#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100000+10

struct node
{
    int l,r,s;
}t[MAXN<<2];
int n,sum;

void build(int l, int r, int num)
{
    t[num].l = l;
    t[num].r = r;
    t[num].s = 1;
    if(l == r) return ;
    int mid = (l+r)>>1;
    build(l, mid, num<<1);
    build(mid+1, r, num<<1|1);
}

void update(int l, int r, int m, int num)
{
    if(t[num].s == m) return ;//如果是相同的，就不用改了
    if(t[num].l == l && t[num].r == r)//
    {
        t[num].s = m;
        return ;
    }
    if(t[num].s != -1)//該區間只有一種顏色
    {
        t[num<<1].s = t[num<<1|1].s = t[num].s;//把他的所有子節點變為父結點一樣的顏色
        t[num].s = -1;//由於該區域顏色與修改不同，所以該區域由純色變為雜色  
    }
    //父區間為雜色對所有子節點進行處理
    int mid = (t[num].l+t[num].r)>>1;
    if(l > mid)
        update(l, r, m, num<<1|1);
    else if(r <= mid)
        update(l, r, m, num<<1);
    else
    {
        update(l, mid, m, num<<1);
        update(mid+1, r, m, num<<1|1);
    }
}

int query(int num)
{
    if(t[num].s != -1)//純色直接找
        return (t[num].r-t[num].l+1)*t[num].s;
    else//不純就找左右子樹
        return query(num<<1)+query(num<<1|1);
}

int main()
{
    int x,y,z,T,k;
    int cas = 1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        build(1, n, 1);
        while(k--)
        {
            scanf("%d%d%d",&x,&y,&z);
            update(x, y, z, 1);
        }
        printf("Case %d: The total value of the hook is %d.\n",cas++,query(1));
    }
    return 0;
}