Count the Colors
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
Description
Your task is counting the segments of different colors you can see at last.
Input
The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.
Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:
x1 x2 c
x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.
All the numbers are in the range [0, 8000], and they are all integers.
Input may contain several data set, process to the end of file.
Output
Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index.
If some color can't be seen, you shouldn't print it.
Print a blank line after every dataset.
Sample Input
5
0 4 4
0 3 1
3 4 2
0 2 2
0 2 3
4
0 1 1
3 4 1
1 3 2
1 3 1
6
0 1 0
1 2 1
2 3 1
1 2 0
2 3 0
1 2 1
Sample Output
1 1
2 1
3 1
1 1
0 2
1 1
线段树区间更新的变形
题意：
在一条长度为8000的线段上染色，每次把区间[a,b]染成c颜色。显然，后面染上去的颜色会覆盖掉之前的颜色。
求染完之后，每个颜色在线段上有多少个间断的区间。
用区间更新的方式，对于区间内的先不更新，当出现新的线段覆盖的时候在pushdown，mark[]遍历一下离根最近的color不为-1（为染色）的就行了，最后通过sum统计出从到大的就好了。
PS：开始的时候由于用来之前的部分代码，结果在pushdown的时候错误成了
tree[tmp].color += tree[x].color;结果出了个Segmentation Fault 从来没有见过的错，结果是ZOJ越界（RE）的错误，查了好久。。。
//线段树区间更新
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define INF 0x3f3f3f3f
#define MAX 8010
#define LL long long
using namespace std;
struct Tree
{
int l,r;
int color;
};
Tree tree[MAX*4];
void pushdown(LL x) ///用于更新color数组
{
LL tmp = x<<1 ;
tree[tmp].color = tree[x].color; ///由子节点通过增加
tree[tmp+1].color = tree[x].color;
tree[x].color=-1;
}
void build(int l,int r,int x)
{
tree[x].l=l , tree[x].r=r , tree[x].color=-1;
if(l==r) return ;
int tmp=x<<1;
int mid=(l+r)>>1;
build(l,mid,tmp);
build(mid+1,r,tmp+1);
}
void update(int l,int r,int c,int x) ///分别表示区间的左 ， 右 ， 增加的值 ，当前父亲节点
{
if(r<tree[x].l||l>tree[x].r) return ;
if(l<=tree[x].l&&r>=tree[x].r) ///该区间为需要更新区间的子区间
{
tree[x].color = c;
return ;
}
if(tree[x].color!=-1) pushdown(x); ///更新从上向下更新color
update(l,r,c,x<<1);
update(l,r,c,(x<<1)+1);
}
int mark[MAX<<2],coun=0; ///注意大小
void query(int l ,int r ,int x )
{
if((l==tree[x].l&&r==tree[x].r && tree[x].color!=-1) || tree[x].l == tree[x].r){
mark[coun++] = tree[x].color; ///要计算的区间包括了该区间
return ;
}
LL tmp=x<<1;
LL mid=(tree[x].l+tree[x].r)>>1;
if(r<=mid) return query(l,r,tmp);
else if(l>mid) return query(l,r,tmp+1);
else{
query(l,mid,tmp) ;
query(mid+1,r,tmp+1);
}
}
int main()
{
int x1[MAX<<2],x2[MAX<<2],c[MAX<<2];
int sum[MAX<<2];
int mmax=-1,m;
while(~scanf("%d",&m))
{
coun = 0;
mmax=-1;
memset(mark,-1,sizeof(mark));
memset(sum,0,sizeof(sum));
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&x1[i],&x2[i],&c[i]);
if(mmax<x1[i]) mmax = x1[i];
if(mmax<x2[i]) mmax = x2[i];
}
build(1,mmax,1);
for( int i = 0;i<m; i++ ){
if(x1[i]+1>x2[i]) continue;
update(x1[i]+1,x2[i],c[i],1);
}
if(mmax<0) continue;
query(1,mmax,1);
for(int i=0;i<coun;){
if(mark[i]==-1){i++ ;continue;}
int x = mark[i];
while(x == mark[++i] && i<coun);
sum[x]++;
}
for(int i=0;i<8010;i++)
if(sum[i])
printf("%d %d\n",i,sum[i]);
printf("\n");
}
return 0;
}
