HDU 1698 Just a Hook (线段树 求和问题)

//4671914 2011-09-28 15:07:14 Accepted 1698 843MS 3260K 1905 B G++ nkhelloworld
//4671915 2011-09-28 15:07:18 Accepted 1698 453MS 3304K 1905 B C++ nkhelloworld
//线段树解决，近似序列求和问题，简单
/*
一个N<=100000个数组成的序列，最多做Q<=100000次操作，每一次操作由a,b,c组成，
意思是将[a,b]区间的值修改为c，求经过Q次操作后序列的总和
*/
#include <cstdio>
#define MAXN 100000
struct SEGMENTTREE
{
int left,right,color;
}tree[MAXN*4+1];

void buildsegtree(int root,int l,int r)
{
tree[root].left = l;    tree[root].right = r;   tree[root].color = 1;
if(l != r)
{
int mid = (l+r)>>1;
buildsegtree(root<<1,l,mid);
buildsegtree(root<<1|1,mid+1,r);
}
}

void update(int root,int a,int b,int c)
{
if(tree[root].left == tree[root].right)
{
tree[root].color = c;
return ;
}
if(tree[root].color != 0)
{
tree[root<<1].color = tree[root].color;
tree[root<<1|1].color = tree[root].color;
tree[root].color = 0;
}
if(a == tree[root].left && b == tree[root].right)
{
tree[root].color = c;
return ;
}
int mid = (tree[root].left + tree[root].right) >>1;
if( b <= mid )
update(root<<1,a,b,c);
else if(a > mid)
update(root<<1|1,a,b,c);
else
{
update(root<<1,a,mid,c);
update(root<<1|1,mid+1,b,c);
}
}

int query(int root,int left,int right)
{
if(tree[root].color!=0)
return (right - left + 1) * tree[root].color;
int mid = (left + right) >> 1;
if(right <= mid)
return query(root<<1, left, right);
else if(left > mid)
return query(root<<1|1, left,right);
else
return query(root<<1, left, mid) + query(root<<1|1, mid+1,right);
}

int main()
{
int totcase,numcase,i,n,q,a,b,c;
scanf("%d",&totcase);
for(numcase = 1;numcase <= totcase; numcase++)
{
scanf("%d%d",&n,&q);
buildsegtree(1,1,n);
for(i=1;i<=q;i++)
{
scanf("%d%d%d",&a,&b,&c);
update(1,a,b,c);
}
printf("Case %d: The total value of the hook is %d.\n",numcase,query(1,1,n));
}
return 0;
}