zoj 1610 Count the Colors（线段树）

#include <stdio.h>
#include <string.h>
#define M 8005
#define L(x) (x<<1)
#define R(x) ((x<<1)+1)

struct data
{
int l,r,col;      //col 等于 -1 表示什么色都没有
}node[M*3];
int color[M];       //记录每块板上的是什么颜色

void BuildTree(int left,int right,int u)
{
node[u].l = left;
node[u].r = right;
node[u].col = -1;
if (left == right)
return ;
int mid = (left + right)>>1;
BuildTree(left,mid,L(u));
BuildTree(mid+1,right,R(u));
}

void getdown(int u)
{
node[L(u)].col = node[u].col;
node[R(u)].col = node[u].col;
node[u].col = -1;
}
void updata (int left,int right,int val,int u)
{
if (node[u].col == val)
return ;
if (left <= node[u].l&&node[u].r <= right)
{
node[u].col = val;
return ;
}
if (node[u].col != -1)       //延迟覆盖 不然可能会TLE（最近做一直都是延迟覆盖）
getdown (u);
int mid = (node[u].l+ node[u].r)>>1;
if (right <= mid)
updata (left,right,val,L(u));
else if (left >= mid+1)
updata (left,right,val,R(u));
else
{
updata (left,right,val,L(u));
updata (left,right,val,R(u));
}

if ((node[L(u)].col == node[R(u)].col)&&node[L(u)].col != -1)  //递归回来修改父结点颜色
node[u].col = node[L(u)].col;

}

void query (int u)
{
if (node[u].col != -1)
{
for (int i = node[u].l;i <= node[u].r;i ++)
color[i] = node[u].col;
return ;
}
if (node[u].l == node[u].r)   //这是不加这个竟然会运行不起来
return ;
query (L(u));
query (R(u));
}
void ans ()
{
int res[M];
memset (res,0,sizeof(res));
int pre = -1,i;
for (i = 0;i < M;i ++)
if (pre != color[i]) //颜色块不相连就++
{
pre = color[i];
res[pre] ++;
}
for (i = 0;i < M;i ++)
if (res[i] !=0)
printf ("%d %d\n",i,res[i]);
printf ("\n");
}
int main ()
{
int n,x,y,val;
while (~scanf ("%d",&n))
{
memset (color,-1,sizeof (color));
BuildTree(0,M,1);
while (n --)
{
scanf ("%d%d%d",&x,&y,&val);
if (x == y)
continue;
updata (x,y-1,val,1);   //后面一个不涂就能避免那种情况
}
query (1);
ans();
}
return 0;
}