zoj 1610 Count the Colors（线段…

题意：给一块长8000米的板上色 问最后能看见几种颜色 而每种颜色的几段

思路：线段树。这道题有很多细节

eg： 0 2 1

3 4 1

output 应该是 1 2 因为中间隔了一段空的

具体见代码

#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;

}