hdu 1698 Just a Hook 線段樹區間更新，hdu1698

hdu 1698 Just a Hook 線段樹區間更新，hdu1698

點擊開啟連結題目連結

Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18010    Accepted Submission(s): 9013


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.

 
看到dota了 - -可惜我是擼狗

金銀銅價值分別為3 2 1

屠夫的鉤子剛開始是銅做的

給出n長度的鉤子 q次操作將a b直接的鉤子替換成價值為c的金屬

問最後價值

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=111111;int sum[MAXN<<2];int lazy[MAXN<<2];void pushup(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt){    lazy[rt]=0;    if(l==r)    {        sum[rt]=1;        return ;    }    int mid=(l+r)>>1;    build(l,mid,rt<<1);    build(mid+1,r,rt<<1|1);    pushup(rt);}void pushdown(int rt,int m){    if(lazy[rt])    {        lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];        sum[rt<<1]=(m-(m>>1))*lazy[rt];        sum[rt<<1|1]=(m>>1)*lazy[rt];        lazy[rt]=0;    }}void update(int L,int R,int c,int l,int r,int rt){    if(l>=L&&R>=r)    {        lazy[rt]=c;        sum[rt]=(r-l+1)*c;        return ;    }    pushdown(rt,r-l+1);    int mid=(l+r)>>1;    if(L<=mid)        update(L,R,c,l,mid,rt<<1);    if(R>mid)        update(L,R,c,mid+1,r,rt<<1|1);    pushup(rt);}int main(){    int t,n,q,a,b,c;    scanf("%d",&t);    for(int cas=1;cas<=t;cas++)    {        scanf("%d %d",&n,&q);        build(1,n,1);        while(q--)        {            scanf("%d %d %d",&a,&b,&c);            update(a,b,c,1,n,1);        }        printf("Case %d: The total value of the hook is %d.\n",cas,sum[1]);    }    return 0;}