Count the Colors
Time Limit: 2 Seconds
Memory Limit: 65536 KB
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
Author: Standlove
Source: ZOJ Monthly, May 2003
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，每种颜色在线段上有多少个间断的区间。
线段树Lazy区间改动。
让我无语的是——输出要求是按颜色编号从0到8000输出的。
而我傻傻的按插入顺序输出了1个小时。。
。o(╯□╰)o
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 8000+10
using namespace std;
int color[MAXN<<2];//对节点染色
void PushDown(int o)
{
if(color[o] != -1)
{
color[o<<1] = color[o<<1|1] = color[o];
color[o] = -1;
}
}
void update(int o, int l, int r, int L, int R, int v)
{
if(L <= l && R >= r)
{
color[o] = v;
return ;
}
PushDown(o);
int mid = (l + r) >> 1;
if(L <= mid)
update(o<<1, l, mid, L, R, v);
if(R > mid)
update(o<<1|1, mid+1, r, L, R, v);
}
int rec[MAXN];//存储i节点的颜色
int top = 0;
void query(int o, int l, int r)
{
if(l == r)
{
rec[top++] = color[o];//记录节点的颜色
return ;
}
int mid = (l + r) >> 1;
PushDown(o);
query(o<<1, l, mid);
query(o<<1|1, mid+1, r);
}
int num[MAXN];//记录颜色出现的 段数
int main()
{
int N;
while(scanf("%d", &N) != EOF)
{
memset(color, -1, sizeof(color));
int a, b, c;
for(int i = 1; i <= N; i++)
{
scanf("%d%d%d", &a, &b, &c);
update(1, 1, 8000, a+1, b, c);//相应颜色数 先加一
}
memset(rec, -1, sizeof(rec));
top = 0;//初始化
query(1, 1, 8000);
memset(num, 0, sizeof(num));//初始化
int i, j;
for(i = 0; i < top;)
{
if(rec[i] == -1)
{
i++;
continue;
}
num[rec[i]]++;
for(j = i + 1; j < top; j++)
{
if(rec[j] != rec[i] || rec[j] == -1)
break;
}
i = j;
}
for(int i = 0; i <= 8000; i++)
{
if(num[i])
printf("%d %d\n", i, num[i]);
}
printf("\n");
}
return 0;
}
直接模拟也能够过：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (10000+10)
#define MAXM (50000000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int num[MAXN];
int color[MAXN];
int main()
{
int n;
while(Ri(n) != EOF)
{
int Max = -INF;
CLR(color, -1);
for(int i = 0; i < n; i++)
{
int x, y, c;
Ri(x); Ri(y); Ri(c);
Max = max(Max, c);
for(int j = x; j < y; j++)
color[j] = c;
}
CLR(num, 0);
for(int i = 0; i <= 8000; )
{
int temp = color[i];
if(temp == -1)
{
i++;
continue;
}
i++;
while(temp == color[i])
i++;
num[temp]++;
}
for(int i = 0; i <= Max; i++)
if(num[i])
printf("%d %d\n", i, num[i]);
printf("\n");
}
return 0;
}