Count the Colors
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
Submit Status Practice ZOJ 1610
Description
Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.
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
错误点：在刚开始以为n就是树的最大范围了，然后一直报段错误（RE），后来发现maxn才是，解决后又wa了几次。还有就是（2,3，0）和（4,5,0）其实是两段，需要注意。然后参考网上的一种写法就过了，至于我的为啥没过，但是别的为啥过了，我也很不解。
解题思路：区间替换、转点为线，统计段数。
/*AC代码*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int maxn=50000;
int col[maxn*4];
int Hash[maxn*2];
int tmp[maxn*2];
int k;
void PushDown(int rt){
if(col[rt]>=0){
col[rt*2]=col[rt];
col[rt*2+1]=col[rt];
col[rt]=-1;
}
}
void update(int rt,int L,int R,int l_ran,int r_ran,int _col){
if(l_ran<=L&&R<=r_ran){
col[rt]=_col;
return ;
}
PushDown(rt);
if(l_ran<=mid){
update(lson,l_ran,r_ran,_col);
}
if(r_ran>mid){
update(rson,l_ran,r_ran,_col);
}
}
void query(int rt,int L,int R){
if(col[rt]>=0){
for(int i=L;i<=R;i++){ //区别在这里
tmp[i]=col[rt];
} //区别在这里
col[rt]=-1;
return ;
}
if(L==R)
return ;
query(lson);
query(rson);
}
void get_ans(int n){
if(tmp[0]>=0)
Hash[tmp[0]]=1;
for(int i=1;i<n;i++){
if(tmp[i]!=tmp[i-1]){
Hash[tmp[i]]++;
}
}
}
void debug2(){
for(int i=0;i<k;i++){
printf("..%d %d\n",i,tmp[i]);
}
}
int main(){
int n;
while(scanf("%d",&n)!=EOF ){
int ta,tb,tc;
memset(tmp,-1,sizeof(tmp));
memset(col,-1,sizeof(col));
for(int i=0;i<n;i++){
scanf("%d%d%d",&ta,&tb,&tc);
update(1,0,maxn,ta*2,tb*2,tc);
}
memset(Hash,0,sizeof(Hash));
k=0;
query(1,0,maxn);
// printf("%d\n",k);
// debug2();
// get_ans(k);
for(int i=0;i<maxn;i++){ //区别在这里
if(tmp[i]!=-1){
int j;
for( j=i;j<maxn&&tmp[i]==tmp[j];j++);
Hash[tmp[i]]++;
i=j-1;
}
} //区别在这里
for(int i=0;i<maxn+20;i++){
if(Hash[i]){
printf("%d %d\n",i,Hash[i]);
}
} printf("\n");
}
return 0;
}
/*WA代码*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int maxn=50000;
int col[maxn*4];
int Hash[maxn*2];
int tmp[maxn*2];
int k=0;
void PushDown(int rt){
if(col[rt]>=0){
col[rt*2]=col[rt];
col[rt*2+1]=col[rt];
col[rt]=-1;
}
}
void update(int rt,int L,int R,int l_ran,int r_ran,int _col){
if(l_ran<=L&&R<=r_ran){
col[rt]=_col;
return ;
}
PushDown(rt);
if(l_ran<=mid){
update(lson,l_ran,r_ran,_col);
}
if(r_ran>mid){
update(rson,l_ran,r_ran,_col);
}
}
void query(int rt,int L,int R){
if(col[rt]>=0){
tmp[k++]=col[rt]; //区别在这里
col[rt]=-1;
return ;
}
if(L==R)
return ;
query(lson);
query(rson);
}
void get_ans(int n){
if(tmp[0]>=0)
Hash[tmp[0]]=1;
for(int i=1;i<n;i++){
if(tmp[i]!=tmp[i-1]){
Hash[tmp[i]]++;
}
}
}
void debug2(){
for(int i=0;i<k;i++){
printf("..%d %d\n",i,tmp[i]);
}
}
int main(){
int n;
while(scanf("%d",&n)!=EOF ){
int ta,tb,tc;
memset(col,-1,sizeof(col));
for(int i=0;i<n;i++){
scanf("%d%d%d",&ta,&tb,&tc);
update(1,0,maxn,ta*2,tb*2,tc);
}
memset(Hash,0,sizeof(Hash));
k=0;
query(1,0,maxn);
// printf("%d\n",k);
// debug2();
get_ans(k); //区别在这里
for(int i=0;i<maxn+20;i++){
if(Hash[i]){
printf("%d %d\n",i,Hash[i]);
}
} printf("\n");
}
return 0;
}