题目链接：https://vjudge.net/problem/ZOJ-1610
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.
<b< dd="">
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.
<b< dd="">
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
<b< dd="">
Sample Output
1 1
2 1
3 1
1 1
0 2
1 1
题解：
问最终用多少段颜色相同的区域。经典的区间染色问题。
写法一：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const double EPS = 1e-8; 15 const int INF = 2e9; 16 const LL LNF = 2e18; 17 const int MAXN = 1e4+10; 18 19 int val[MAXN<<2]; 20 int num[MAXN], color[MAXN]; 21 22 void push_down(int u) 23 { 24 if(val[u]>=0) 25 { 26 val[u*2] = val[u*2+1] = val[u]; 27 val[u] = -1; 28 } 29 } 30 31 void set_val(int u, int l, int r, int x, int y, int _val) 32 { 33 if(x<=l && r<=y) 34 { 35 val[u] = _val; 36 return; 37 } 38 39 push_down(u); 40 int mid = (l+r)>>1; 41 if(x<=mid) set_val(u*2, l, mid, x, y, _val); 42 if(y>=mid+1) set_val(u*2+1, mid+1, r, x, y, _val); 43 } 44 45 void query(int u, int l, int r) 46 { 47 if(l==r) 48 { 49 color[l] = val[u]; 50 return; 51 } 52 53 push_down(u); 54 int mid = (l+r)>>1; 55 query(u*2, l, mid); 56 query(u*2+1, mid+1, r); 57 } 58 59 int main() 60 { 61 int m; 62 while(scanf("%d", &m)!=EOF) 63 { 64 memset(val, -1, sizeof(val)); 65 for(int i = 1; i<=m; i++) 66 { 67 int x, y, z; 68 scanf("%d%d%d", &x, &y, &z); 69 if(x<y) set_val(1, 1, 8000, x+1, y, z); 70 } 71 72 query(1, 1, 8000); 73 memset(num, 0, sizeof(num)); 74 for(int i = 1; i<=8000; i++) 75 if(color[i]!=-1 && (i==1 || color[i]!=color[i-1])) 76 num[color[i]]++; 77 78 79 for(int i = 0; i<=8000; i++) 80 if(num[i]) 81 printf("%d %d ", i, num[i]); 82 printf(" "); 83 } 84 }
写法二：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const double EPS = 1e-8; 15 const int INF = 2e9; 16 const LL LNF = 2e18; 17 const int MAXN = 1e4+10; 18 19 int val[MAXN<<2]; 20 int num[MAXN]; 21 22 void push_down(int u) 23 { 24 if(val[u]>=0) 25 { 26 val[u*2] = val[u*2+1] = val[u]; 27 val[u] = -1; 28 } 29 } 30 31 void set_val(int u, int l, int r, int x, int y, int _val) 32 { 33 if(x<=l && r<=y) 34 { 35 val[u] = _val; 36 return; 37 } 38 39 push_down(u); 40 int mid = (l+r)>>1; 41 if(x<=mid) set_val(u*2, l, mid, x, y, _val); 42 if(y>=mid+1) set_val(u*2+1, mid+1, r, x, y, _val); 43 } 44 45 int pre; 46 void query(int u, int l, int r) 47 { 48 if(l==r) 49 { 50 if(val[u]>=0 && val[u]!=pre) 51 num[val[u]]++; 52 pre = val[u]; 53 return; 54 } 55 56 push_down(u); 57 int mid = (l+r)>>1; 58 query(u*2, l, mid); 59 query(u*2+1, mid+1, r); 60 } 61 62 int main() 63 { 64 int m; 65 while(scanf("%d", &m)!=EOF) 66 { 67 memset(val, -1, sizeof(val)); 68 for(int i = 1; i<=m; i++) 69 { 70 int x, y, z; 71 scanf("%d%d%d", &x, &y, &z); 72 if(x<y) set_val(1, 1, 8000, x+1, y, z); 73 } 74 75 memset(num, 0, sizeof(num)); 76 pre = -1; 77 query(1, 1, 8000); 78 for(int i = 0; i<=8000; i++) 79 if(num[i]) 80 printf("%d %d ", i, num[i]); 81 printf(" "); 82 } 83 }