ZOJ - 1610 Count the Colors 线段树区间修改

Posted 2020-10-21 晓风微微

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

 

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Author: Standlove
Source: ZOJ Monthly, May 2003

题意：

给出l,r点的区间，给这些点的区间染上颜色c

问最后每种颜色有几块(也就是有几个不连续的区间块)

输出按颜色的标号从小到大输出

输出内容为

每行两个整数

第一个整数为颜色标号

第二个整数位该颜色区间块的个数

做法：

和

http://poj.org/problem?id=2528

一样的意思，只不过把区间修改变成点修改，然后去掉离散化，再加上一个统计连续块就好

代码

  1 #include<iostream>
  2 using namespace std;
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<map>
  6 #include<algorithm>
  7 const int MAXN=32000;
  8 typedef int LL;
  9 int v[MAXN*3];
 10 int ans[MAXN*3];
 11 int zans[MAXN*3];
 12 inline int read()
 13 {
 14     int k=0;
 15     char f=1;
 16     char c=getchar();
 17     for(;!isdigit(c);c=getchar() )
 18         if(c==‘-‘)
 19             f=-1;
 20     for(;isdigit(c);c=getchar() )
 21         k=k*10+c-‘0‘;
 22     return k*f;
 23 }
 24 template<class T>
 25 class tree{
 26     public:
 27     int ws;
 28     T a[150000];
 29     T gg[150000];
 30     int l[150000],r[150000];
 31     void creat(int k,int ll,int rr){
 32         l[k]=ll;r[k]=rr;
 33         if(ll>=rr){
 34             a[k]=-1;
 35             gg[k]=-1;
 36             return;
 37         }
 38         int mid=(ll+rr)/2; 
 39         creat(k*2,ll,mid);
 40         creat(k*2+1,mid+1,rr);
 41         a[k]=a[k*2]+a[k*2+1];
 42         gg[k]=-1;
 43     }
 44     void change(int k,int ll,int rr,int t){
 45         if(gg[k]!=-1){
 46             a[k]=gg[k]*(r[k]-l[k]+1);
 47             if(l[k]!=r[k]){
 48                 gg[k*2]=gg[k];
 49                 gg[k*2+1]=gg[k];
 50             }
 51             gg[k]=-1;
 52         }
 53         if(ll>r[k]||rr<l[k]) return;
 54         if(ll<=l[k]&&rr>=r[k]){
 55             a[k]=(r[k]-l[k]+1)*t;
 56             if(l[k]!=r[k]){
 57                 gg[k*2]=t;
 58                 gg[k*2+1]=t;
 59             }
 60             return;
 61         }
 62         change(k*2,ll,rr,t);
 63         change(k*2+1,ll,rr,t);
 64         a[k]=a[k*2]+a[k*2+1];
 65     }
 66     LL query(int k,int ll,int rr){
 67         if(gg[k]!=-1){
 68             a[k]=(r[k]-l[k]+1)*gg[k];
 69             if(l[k]!=r[k]){
 70                 gg[k*2]=gg[k];
 71                 gg[k*2+1]=gg[k];
 72             }
 73             gg[k]=-1;
 74         }
 75         if(ll>r[k]||rr<l[k]) return 0; 
 76         if(l[k]==r[k]){
 77             return 0;
 78         }
 79         int ans1=query(k*2,ll,rr);
 80         int ans2=query(k*2+1,ll,rr);
 81         return 0;    
 82     }
 83     void get_ans(int k){
 84         if(l[k]==r[k]){
 85         //    cout<<"data: "<<l[k]<<" "<<a[k]<<endl;
 86             zans[l[k]]=a[k];
 87             if(a[k]<0)
 88                 return;
 89             if(v[a[k]])
 90                 return;
 91             else{
 92                 v[a[k]]=1;
 93                 return;
 94             }
 95         }
 96         get_ans(2*k);
 97         get_ans(2*k+1);
 98     }
 99 };
100 tree<int> line_tree;
101 int a[MAXN*3];
102 int b[MAXN*3];
103 int deal(){
104     int n;
105     if(scanf("%d",&n)==EOF)
106         return 1;
107     memset(zans,0,sizeof(zans));
108     int l,r;
109     int maxn,minn;
110     maxn=-1;
111     minn=0x7fffffff;
112     for(int i=0;i<n;i++){
113         a[i*2]=read();//左区间 
114         a[i*2+1]=read();//右区间 
115         if(maxn<a[i*2])
116             maxn=a[i*2];
117         if(maxn<a[i*2+1])
118             maxn=a[i*2+1];
119         if(minn>a[i*2])
120             minn=a[i*2];
121         if(minn>a[i*2+1])
122             minn=a[i*2+1];
123         b[i]=read();//染色颜色 
124     }
125     memset(v,0,sizeof(v));
126     line_tree.creat(1,minn,maxn);
127     for(int i=0;i<n;i++){
128         if(a[i*2]>a[i*2+1])
129             swap(a[i*2],a[i*2+1]);
130         line_tree.change(1,a[i*2],a[i*2+1]-1,b[i]);
131     }
132     line_tree.query(1,minn,maxn);
133     line_tree.get_ans(1);
134     memset(ans,0,sizeof(ans));
135     for(int i=minn;i<=maxn;i++){
136         if((i==minn||zans[i]!=zans[i-1])&&v[zans[i]])
137             ans[zans[i]]++;
138     }
139     for(int i=0;i<=8010;i++){
140         if(v[i]==0)
141             continue;
142         else
143         printf("%d %d\n",i,ans[i]);
144     }
145     return 0;
146 }
147 int main(){
148     while(1){
149         if (deal())
150             break;
151         putchar(‘\n‘);
152     }
153     return 0;
154 }