ZJU 1610 - Count the Colors (线段树 + 区间修改)
代码
基础题
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <ctime>
#include <cstdlib>
#include <fstream>
#include <string>
#include <sstream>
#include <map>
#include <cmath>
#define LL long long
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 8000 + 5;
const int MOD = 20071027;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
int col[MAXN << 2], cnt, vis[MAXN];
map<int, int> mp;
void PushDown(int rt)
{
if (col[rt] != -1)
{
col[LRT] = col[RRT] = col[rt];
col[rt] = -1;
return;
}
}
void Query(int rt, int l, int r)
{
if (col[rt] != -1)
{
fill(vis + l, vis + r + 1, col[rt]);
return;
}
if (l == r) return;
int mid = MID(l, r);
Query(LC); Query(RC);
}
void Update(int rt, int l, int r, int L, int R, int val)
{
if (L <= l && r <= R)
{
col[rt] = val;
return;
}
PushDown(rt);
int mid = MID(l, r);
if (L <= mid) Update(LC, L, R, val);
if (R > mid) Update(RC, L, R, val);
}
int main()
{
//ROP;
int n, i, j;
while (~scanf("%d", &n))
{
cnt = 0;
MS(col, -1), MS(vis, -1);
mp.clear();
int a, b, c;
for (i = 0; i < n; i++)
{
scanf("%d%d%d", &a, &b, &c);
Update(1, 1, 8000, a + 1, b, c);
}
Query(1, 1, 8000);
for (i = 1; i < MAXN; i++)
{
if (vis[i] != -1)
{
for (j = i; j <= MAXN && vis[j] == vis[i]; j++);
mp[vis[i]]++;
i = j - 1;
}
}
for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++)
printf("%d %d\n", it->F, it->S);
puts("");
}
return 0;
}