hdu 1698 Just a Hook（線段樹之 成段更新）

標籤：線段樹

Just a Hook                                                                            Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem DescriptionIn 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.
 
InputThe 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.
 
OutputFor 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

11021 5 25 9 3

 
Sample Output

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

 題意：有一個N段金屬組成的鉤子，開始時這N段全部是cupreous，接下來有Q次操作，每次操作為x,y,z，表示把x到y這一段換成z，z為1時為cupreous，價值為1；z為2時為silver，價值為2；z為3時為golden，價值為3；問這個鉤子最後的總價值是多少。
分析：線段樹成段更新，每次記錄區間和，最後輸出根節點的sum即可。

#include<cstdio>#define lson l, mid, root<<1#define rson mid+1, r, root<<1|1const int N = 100010;struct node{    int l;    int r;    int sum;    int color;}a[N<<2];void PushUp(int root){    a[root].sum = a[root<<1].sum + a[root<<1|1].sum;}void PushDown(int len, int root){    if(a[root].color)    {        a[root<<1].color = a[root<<1|1].color = a[root].color;        a[root<<1].sum = (len - (len>>1)) * a[root].color;        a[root<<1|1].sum = (len>>1) * a[root].color;        a[root].color = 0;    }}void build_tree(int l, int r, int root){    a[root].l = l;    a[root].r = r;    a[root].color = 0;    if(l == r)    {        a[root].sum = 1;        return ;    }    int mid = (l + r) >> 1;    build_tree(lson);    build_tree(rson);    PushUp(root);}void update(int l, int r, int root, int z){    if(l <= a[root].l && r >= a[root].r)    {        a[root].color = z;        a[root].sum = (a[root].r - a[root].l + 1) * z;        return;    }    PushDown(a[root].r - a[root].l + 1, root);    int mid = (a[root].l + a[root].r) >> 1;    if(l <= mid) update(l, r, root<<1, z);    if(r > mid) update(l, r, root<<1|1, z);    PushUp(root);}int main(){    int T, n, m, cas = 0;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        build_tree(1, n, 1);        scanf("%d",&m);        int x, y, z;        while(m--)        {            scanf("%d%d%d",&x,&y,&z);            update(x, y, 1, z);        }        int ans = a[1].sum;        printf("Case %d: The total value of the hook is %d.\n", ++cas, ans);    }    return 0;}