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  • HDU-1698 Just A Hook(线段树)

    Just a Hook

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 29254    Accepted Submission(s): 14462


    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
     
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    嗯线段树区间查询区间修改嗯
     1 #include "bits/stdc++.h"
     2 #define lson rt<<1,l,m
     3 #define rson rt<<1|1,m+1,r
     4 using namespace std;
     5 typedef long long LL;
     6 const int MAX=100005;
     7 int cas,n,m;
     8 int c[MAX<<2],t[MAX<<2];
     9 void PushUp(int rt){
    10     c[rt]=c[rt<<1]+c[rt<<1|1];
    11 }
    12 void PushDown(int rt,int m){
    13     if (t[rt]){
    14         t[rt<<1]=t[rt<<1|1]=t[rt];
    15         c[rt<<1]=t[rt<<1]*(m-(m>>1));
    16         c[rt<<1|1]=t[rt<<1]*(m>>1);
    17         t[rt]=0;
    18     }
    19 }
    20 void build(int rt,int l,int r){
    21     t[rt]=1;
    22     c[rt]=(r-l+1);
    23     if (l==r) return;
    24     int m=(l+r)>>1;
    25     build(lson);
    26     build(rson);
    27 }
    28 void update(int rt,int l,int r,int x,int y,int z){
    29     if (x<=l && r<=y){
    30         t[rt]=z;
    31         c[rt]=t[rt]*(r-l+1);
    32         return;
    33     }
    34     int m=(l+r)>>1;
    35     PushDown(rt,r-l+1);
    36     if (x<=m)
    37      update(lson,x,y,z);
    38     if (y>m)
    39      update(rson,x,y,z);
    40     PushUp(rt);
    41 }
    42 int main(){
    43     freopen ("hook.in","r",stdin);
    44     freopen ("hook.out","w",stdout);
    45     int i,j,T;
    46     int x,y,z;
    47     scanf("%d",&T);
    48     while (T--){
    49         scanf("%d",&n);
    50         build(1,1,n);
    51         scanf("%d",&m);
    52         for (i=1;i<=m;i++){
    53             scanf("%d%d%d",&x,&y,&z);
    54             update(1,1,n,x,y,z);
    55         }
    56         printf("Case %d: The total value of the hook is %d.
    ",++cas,c[1]);
    57     }
    58     return 0;
    59 }
    我已无力吐槽
    未来是什么样,未来会发生什么,谁也不知道。 但是我知道, 起码从今天开始努力, 肯定比从明天开始努力, 要快一天实现梦想。 千里之行,始于足下! ——《那年那兔那些事儿》
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  • 原文地址:https://www.cnblogs.com/keximeiruguo/p/6068140.html
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