ZOJ 1610 Count the Colors （线段树区间染色）

题意: 不断对区间染色 最后求每种颜色有连续的几段 按字典序输出

分析: 由于是区间染色啦 假如将区间全部看成点

          [2,3] 1    [3,4] 2 这样的话 本来3颜色是1 却变成了2 会有问题

          解决方式将[l,r]区间更新的时候用[l,r-1]来代替 保持原来的区间的性质

          问题就 转化成了线段树区间更新的问题了

代码:

//
// Created by TaoSama on 2015-09-16
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 8e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, lastColor;
int row[N << 2];
map<int, int> mp;

#define root 1, n, 1
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

void push_down(int rt) {
if(~row[rt]) {
row[rt << 1] = row[rt << 1 | 1] = row[rt];
row[rt] = -1;
}
}

void update(int L, int R, int c, int l, int r, int rt) {
if(L <= l && r <= R) {
row[rt] = c;
return;
}
int m = l + r >> 1;
push_down(rt);
if(L <= m) update(L, R, c, lson);
if(R > m) update(L, R, c, rson);
}

void query(int l, int r, int rt) {
if(l == r) {
if(row[rt] != lastColor) ++mp[row[rt]];
lastColor = row[rt];
return;
}
int m = l + r >> 1;
push_down(rt);
query(lson);
query(rson);
}

int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d", &n) == 1) {
memset(row, -1, sizeof row);
for(int i = 1; i <= n; ++i) {
int x, y, c; scanf("%d%d%d", &x, &y, &c);
update(x, y - 1, c, 0, 8000, 1);
}

lastColor = -1; mp.clear();
query(0, 8000, 1);

for(map<int, int>::iterator iter = mp.begin(); iter != mp.end(); ++iter) {
if(iter->first == -1) continue;
printf("%d %d\n", iter->first, iter->second);
}
puts("");
}
return 0;
}