hdoj 1698 Just a Hook【线段树区间修改】

Just a Hook

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23137    Accepted Submission(s):
11600

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.

 题意：输入t代表有t组测试数据，然后一个数字n代表n个英雄，接下来数字q代表接下来有q行，每行三个整数a，b，c，代表将区间a到b的值全部换为c，最后问总分数，注意  初始分数为1

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define MAX 100100
#define INF 0x3f3f3f
using namespace std;
int sum[MAX<<2];
int change[MAX<<2];
void pushup(int o)
{
	sum[o]=sum[o<<1]+sum[o<<1|1];
}
void pushdown(int o,int m)
{
	if(change[o])
	{
		change[o<<1]=change[o<<1|1]=change[o];
		sum[o<<1]=change[o]*(m-(m>>1));
		sum[o<<1|1]=change[o]*(m>>1);
		change[o]=0;
	}
}
void gettree(int o,int l,int r)
{
	sum[o]=1;change[o]=0;
	if(l==r)
	    return ;
	int mid=(l+r)>>1;
	gettree(o<<1,l,mid);
	gettree(o<<1|1,mid+1,r);
	pushup(o);
}
void update(int o,int l,int r,int L,int R,int v)
{
	if(L<=l&&R>=r)
	{
		change[o]=v;
		sum[o]=v*(r-l+1);
		return ;
	}
	pushdown(o,r-l+1);
	int mid=(r+l)>>1;
	if(L<=mid)
	    update(o<<1,l,mid,L,R,v);
	if(R>mid)
	    update(o<<1|1,mid+1,r,L,R,v);
    pushup(o);
}
int find(int o,int l,int r,int L,int R)
{
	if(L<=l&&R>=r)
	{
		return sum[o];
	}
	pushdown(o,r-l+1);
	int ans=0;
	int mid=(r+l)>>1;
	if(L<=mid)
	    ans+=find(o<<1,l,mid,L,R);
	if(R>mid)
	    ans+=find(o<<1|1,mid+1,r,L,R);
	return ans;
}
int main()
{
	int t,k,i,j;
	int n,m;
	k=1;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		int a,b,c;
		gettree(1,1,n);
		while(m--)
		{
			scanf("%d%d%d",&a,&b,&c);
			update(1,1,n,a,b,c);
		}
		printf("Case %d: The total value of the hook is ",k++);
		printf("%d.\n",find(1,1,n,1,n));
	}
	return 0;
}