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): 24473 Accepted Submission(s): 12193

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
2008 “Sunline Cup” National Invitational Contest

题意：一段线段由n条小线段组成，每次操作把一个区间的小线段变成金银铜之一（金的价值为3，银为2，铜为1），最初可当做全为铜；最后求这条线段的总价值。
比较基础的区间更新，但是比较丧心病狂的是末尾还有个 ‘.’ ！

#include<bits/stdc++.h>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define maxn 111111long long tree[maxn<<2];long long flag[maxn<<2];void push_up(long long rt){    tree[rt]=tree[rt<<1]+tree[rt<<1|1];}void push_down(long long rt,long long m){    if(flag[rt])    {        flag[rt<<1]=flag[rt<<1|1]=flag[rt];        tree[rt<<1]=flag[rt]*(m-(m>>1));        tree[rt<<1|1]=flag[rt]*(m>>1);        flag[rt]=0;    }}void build(long long l,long long r,long long rt){    flag[rt]=0;    if(l==r)        tree[rt]=1;    else {        long long m=(l+r)>>1;        build(lson);        build(rson);        push_up(rt);    }}void update(long long L,long long R,long long add,long long l,long long r,long long rt){    if(L<=l&&r<=R)    {        tree[rt]=add*(r-l+1);        flag[rt]=add;        return;    }    else {        push_down(rt,r-l+1);        long long m=(l+r)>>1;        if(L<=m) update(L,R,add,lson);        if(m<R) update(L,R,add,rson);        push_up(rt);    }}int main(){    long long t,m,n;    cin>>t;    for(long long ci=1;ci<=t;ci++)    {        scanf("%lld%lld",&n,&m);        build(1,n,1);        long long a,b,c;        for(long long i=0;i<m;i++)        {            scanf("%lld%lld%lld",&a,&b,&c);            update(a,b,c,1,n,1);        }        printf("Case %d: The total value of the hook is ",ci);        cout<<tree[1]<<'.'<<endl;    }}