hdu1698-Just a Hook(线段树)

转换成区间染色问题吧。。。。

给你一个区间，默认是第一种色，然后再给出一个区间以及一种颜色开始染，最后是求出整个区间上所有颜色的价值是多少，三种色，1，2，3

类似于区间求和问题，常规不能用依旧需要保持一个增量来做，还有个办法就是区间染色问题来做，维持区间的颜色和不是和，最后get把每个区间段的颜色求出

最后再求个和就ok...

区间求和

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
#include <climits>
#define L(x) (x << 1)
#define R(x) (x << 1 | 1)

using namespace std;

const int MAX = 100010;
struct Tnode{ int val,a; int l,r;};
Tnode node[MAX*4];
void init()
{
	memset(node,0,sizeof(node));
}
void build(int t,int l,int r)
{
	node[t].l = l;
	node[t].r = r;
	if(l==r-1)
	{
	    node[t].val = 1;
	return ;
	}
	int mid = ( l + r ) >> 1;
	build(L(t), l, mid);
	build(R(t), mid, r);
	node[t].val=node[L(t)].val+node[R(t)].val;
}

void update(int t,int l,int r,int val)
{
	if( node[t].l==l && node[t].r==r )
	{
	    node[t].a+=val;
		node[t].val = val*(r-l);
		return ;
	}
	if(node[t].a)
	{
	    node[L(t)].a+=node[t].a;
	    node[R(t)].a+=node[t].a;
	    node[R(t)].val += node[t].a*(node[R(t)].r - node[R(t)].l);
		node[L(t)].val += node[t].a*(node[L(t)].r - node[L(t)].l);
	    node[t].a=0;
	}
	int mid = (node[t].l + node[t].r) >> 1;
	if( l >=mid )
		update(R(t),l,r,val);
	else if( r <= mid )
        update(L(t),l,r,val);
        else
        {
            update(R(t),mid,r,val);
            update(L(t),l,mid,val);
        }
    node[t].val=node[L(t)].val+node[R(t)].val;
}

int get(int t,int l,int r)
{
	if( node[t].l == l && node[t].r ==r )
	{
		return node[t].val;
	}
	if(node[t].a)
	{
	    node[L(t)].a+=node[t].a;
	    node[R(t)].a+=node[t].a;
	    node[R(t)].val += node[t].a*(node[R(t)].r - node[R(t)].l);
		node[L(t)].val += node[t].a*(node[L(t)].r - node[L(t)].l);
	    node[t].a=0;
	}
	int mid = (node[t].l + node[t].r) >> 1;
	if( l >= mid )
		return get(R(t),l,r);
    else if( r <= mid )
        return get(L(t),l,r);
        else
        {
           return get(R(t),mid,r)+get(L(t),l,mid);
        }
}

int main()
{
	int n,m,x,y,val,ind=1;
	char str[5];
	int T;
	scanf("%d",&T);
	while(T-- )
	{
		init();
		scanf("%d%d",&n,&m);
		build(1,1,n+1);
		while( m-- )
		{
			scanf("%d%d%d",&x,&y,&val);
			update(1,x,y+1,val);
		}
		printf("Case %d: The total value of the hook is %d.\n",ind++,get(1,1,n+1));
	}

return 0;
}

区间染色

/*Author:zxy_snow*/
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
#include <climits>
#define MID(x,y) ((x+y)>>1)
#define L(x) (x<<1)
#define R(x) (x<<1|1)

using namespace std;

const int MAX = 100010;
struct Tnode{int sum,l,r;};
Tnode node[MAX*4];
void init()
{
	memset(node,0,sizeof(node));
}
void Build(int t,int l,int r)
{
	node[t].sum = 1;
	node[t].l = l; node[t].r = r;
	if( l == r - 1 ) return ;
	int mid = MID(l,r);
	Build(R(t),mid,r);
	Build(L(t),l,mid);
}

void Updata(int t,int l,int r,int sum)
{
	if( node[t].l == l && node[t].r == r )
	{
		node[t].sum = sum;
		return ;
	}
	if( node[t].sum > 0 )
	{
		node[R(t)].sum = node[t].sum;
		node[L(t)].sum = node[t].sum;
		node[t].sum = -1;
	}
	int mid = MID(node[t].l,node[t].r);
	if( l >= mid )
		Updata(R(t),l,r,sum);
	else
		if( r <= mid )
			Updata(L(t),l,r,sum);
		else
		{
			Updata(L(t),l,mid,sum);
			Updata(R(t),mid,r,sum);
		}
}

int Query(int t,int l,int r)
{
	if( node[t].sum > 0 )
		return node[t].sum * (r - l);
	int mid = MID(node[t].l,node[t].r);
	if( l >= mid )
		return Query(R(t),l,r);
	else
		if( r <= mid )
			return Query(L(t),l,r);
		else
			return Query(L(t),l,mid) + Query(R(t),mid,r);
}
int main()
{
	int ncases,ind = 1,n,m,x,y,z;
	
	scanf("%d",&ncases);
	
	while( ncases-- )
	{
		scanf("%d%d",&n,&m);
		init();
		Build(1,0,n);
		while( m-- )
		{
			scanf("%d%d%d",&x,&y,&z);
			Updata(1,x-1,y,z);
		}
		printf("Case %d: The total value of the hook is %d.\n",ind++,Query(1,0,n));
	}

return 0;
}