Problem A hdu 1698 Just a Hook

Just a Hook

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

Total Submission(s): 11857 Accepted Submission(s): 5890

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.

Source2008 “Sunline Cup” National Invitational Contest

Recommendwangyeps：感謝博博的解題報告，線段樹是博博的拿手好戲。

題目大意。玩過dota遊戲的朋友應該知道這樣的一個英雄。該題對他的武器進行討論。它的武器是一條長長的鏈子。由不同價值的材料構成。現對它的鏈子進行q次操作。每次把[x,y]間的鏈節價值改為z。為q次操作後鏈子的總價值

#include <stdio.h>struct node//定義線段樹區間結構{    int l,r,v,sum,en;//l,r分別代表左右端點。v鏈子的價值，sum為[l,r]鏈子的總價值。                     //en標記[l,r]鏈子價值是否相同} sbt[400010];//4倍就合適了。void btree(int l,int r,int k)//以[l,r]建立線段樹.k為結點號下同{    int mid,ls,rs;//mid記錄線段中點。ls,rs左右兒子    sbt[k].l=l;    sbt[k].r=r;    sbt[k].v=1;//價值初始化為1，    sbt[k].en=1;//初始化為相同    if(l==r)//如果到了葉子結點。初始sum    {        sbt[k].sum=1;        return;    }    ls=k<<1;//計算左右兒子標號    rs=k<<1|1;    mid=(l+r)>>1;    btree(l,mid,ls);//遞迴構造線段樹    btree(mid+1,r,rs);    sbt[k].sum=sbt[ls].sum+sbt[rs].sum;//遞迴計算區間[l,r]的總價值}void update(int l,int r,int k,int d)//將[l,r]的價值置為d{    int mid,ls,rs;//同上    if(sbt[k].l==l&&sbt[k].r==r)//如果區間匹配。直接標記返回。    {        sbt[k].en=1;        sbt[k].v=d;//更新價值        sbt[k].sum=(sbt[k].r-sbt[k].l+1)*d;//計算總價值        return;    }    ls=k<<1;    rs=k<<1|1;    mid=(sbt[k].l+sbt[k].r)>>1;    if(sbt[k].en)//如果修改前區間價值相同    {        sbt[k].en=0;//取消標幟        sbt[ls].en=sbt[rs].en=1;//標誌下傳        sbt[ls].v=sbt[rs].v=sbt[k].v;//更新價值        sbt[k].v=0;        sbt[ls].sum=(sbt[ls].r-sbt[ls].l+1)*sbt[ls].v;//更新左右兒子總價值        sbt[rs].sum=(sbt[rs].r-sbt[rs].l+1)*sbt[rs].v;    }    if(r<=mid)//遞迴更新        update(l,r,ls,d);    else if(l>mid)        update(l,r,rs,d);    else    {        update(l,mid,ls,d);        update(mid+1,r,rs,d);    }    sbt[k].sum=sbt[ls].sum+sbt[rs].sum;//遞迴計算更新後區間[l,r]的總價值}int qu(int l,int r,int k)//詢問[l,r]間的總價值{    int mid,ls,rs;    if(sbt[k].en||(sbt[k].l==l&&sbt[k].r==r))//如果被詢問區間剛好匹配或價值相同       return sbt[k].sum;//直接返回總價值    ls=k<<1;    rs=k<<1|1;    mid=(l+r)>>1;    if(l>mid)//遞迴詢問        return qu(l,r,rs);    else if(r<=mid)        return qu(l,r,ls);    else        return qu(l,mid,ls)+qu(mid+1,r,rs);}int main(){    int t,x,y,z,n,q,i,j;//如題意    scanf("%d",&t);    for(i=1;i<=t;i++)    {        scanf("%d",&n);        btree(1,n,1);        scanf("%d",&q);        for(j=1;j<=q;j++)        {            scanf("%d%d%d",&x,&y,&z);            update(x,y,1,z);        }        printf("Case %d: The total value of the hook is %d.\n",i,qu(1,n,1));    }    return 0;}