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 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

11021 5 25 9 3

Sample Output

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

Source

2008 “Sunline Cup” National Invitational Contest

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wangye

题目大意。玩过dota游戏的朋友应该知道这样的一个英雄。该题对他的武器进行讨论。它的武器是一条长长的链子。由不同价值的材料构成。现对它的链子进行q次操作。每次把[x,y]间的链节价值改为z。为q次操作后链子的总价值

#include <stdio.h>struct node//定义线段树区间结构{    int l,r,v,sum,en;//l,r分别代表左右端点。v链子的价值，sum为[l,r]链子的总价值。                     //en标记[l,r]链子价值是否相同} sbt[400010];//4倍就合适了。void btree(int l,int r,int k)//以[l,r]建立线段树.k为结点号下同{    int mid,ls,rs;//mid记录线段中点。ls,rs左右儿子    sbt[k].l=l;    sbt[k].r=r;    sbt[k].v=1;//价值初始化为1，    sbt[k].en=1;//初始化为相同    if(l==r)//如果到了叶子结点。初始sum    {        sbt[k].sum=1;        return;    }    ls=k<<1;//计算左右儿子标号    rs=k<<1|1;    mid=(l+r)>>1;    btree(l,mid,ls);//递归构造线段树    btree(mid+1,r,rs);    sbt[k].sum=sbt[ls].sum+sbt[rs].sum;//递归计算区间[l,r]的总价值}void update(int l,int r,int k,int d)//将[l,r]的价值置为d{    int mid,ls,rs;//同上    if(sbt[k].l==l&&sbt[k].r==r)//如果区间匹配。直接标记返回。    {        sbt[k].en=1;        sbt[k].v=d;//更新价值        sbt[k].sum=(sbt[k].r-sbt[k].l+1)*d;//计算总价值        return;    }    ls=k<<1;    rs=k<<1|1;    mid=(sbt[k].l+sbt[k].r)>>1;    if(sbt[k].en)//如果修改前区间价值相同    {        sbt[k].en=0;//取消标记        sbt[ls].en=sbt[rs].en=1;//标志下传        sbt[ls].v=sbt[rs].v=sbt[k].v;//更新价值        sbt[k].v=0;        sbt[ls].sum=(sbt[ls].r-sbt[ls].l+1)*sbt[ls].v;//更新左右儿子总价值        sbt[rs].sum=(sbt[rs].r-sbt[rs].l+1)*sbt[rs].v;    }    if(r<=mid)//递归更新        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;//递归计算更新后区间[l,r]的总价值}int qu(int l,int r,int k)//询问[l,r]间的总价值{    int mid,ls,rs;    if(sbt[k].en||(sbt[k].l==l&&sbt[k].r==r))//如果被询问区间刚好匹配或价值相同       return sbt[k].sum;//直接返回总价值    ls=k<<1;    rs=k<<1|1;    mid=(sbt[k].l+sbt[k].r)>>1;    if(l>mid)//递归询问        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;//如题意    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;}