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ZOJ1610 Count the Colors 经典线段树染色问题

题意,给你n个  x,y,c,意思就是区间[x,y]被染成C色,但是颜色会被覆盖的,染色操作完成以后 问你每种颜色有多少段 并输出颜色编号id跟段数cnt


经典问题,不过写的有点撮吧,没去看别人的,这个方法应该是最传统的最普通的,常规的开数组记录,也许大神们有更高端的方法



#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>
#include<cctype>

#define ll long long

#define LL __int64

#define eps 1e-8

#define inf 0xfffffff

//const LL INF = 1LL<<61;

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int > P;
//vector<pair<int,int> > ::iterator iter;
//
//map<ll,int >mp;
//map<ll,int >::iterator p;

const int N = 8000 + 5;

typedef struct Node {
	int l,r;
	int col;
};

Node tree[N * 4];

int color[N];
int ans[N];

void init() {
	memset(color,-1,sizeof(color));
	memset(ans,0,sizeof(ans));
}

void build(int l,int r,int id) {
	tree[id].l = l;
	tree[id].r = r;
	tree[id].col = -1;
	if(l == r)return;
	int mid = (l + r)/2;
	build(l,mid,id<<1);
	build(mid+1,r,id<<1|1);
}

void cal(int id) {
	if(tree[id].col > -1) {
		tree[id<<1].col = tree[id].col;
		tree[id<<1|1].col = tree[id].col;
		tree[id].col = -1;
	}		
}

void update(int l,int r,int id,int col) {
	if(tree[id].col == col)return;
	if(l <= tree[id].l && r >= tree[id].r) {
		 tree[id].col = col;return;
	}
	cal(id);
	int mid = (tree[id].l + tree[id].r)/2;
	if(r <= mid)update(l,r,id<<1,col);
	else if(l > mid)update(l,r,id<<1|1,col);
	else {
		update(l,mid,id<<1,col);
		update(mid+1,r,id<<1|1,col);
	}
}

void query(int l,int r,int id) {
	if(tree[id].col > -1) {
		for(int i=tree[id].l;i<=tree[id].r;i++)
			color[i] = tree[id].col;
		return;
	}
	if(tree[id].l == tree[id].r)return;
	int mid = (tree[id].l + tree[id].r)/2;
	if(r <= mid)query(l,r,id<<1);
	else if(l > mid)query(l,r,id<<1|1);
	else {
		query(l,mid,id<<1);
		query(mid+1,r,id<<1|1);
	}
}

int main() {
	int n;
	while(scanf("%d",&n) == 1) {
		int q = n;
		init();
		build(1,N,1);
		int x,y,c;
		while(q--) {
			scanf("%d %d %d",&x,&y,&c);
			x++;
			update(x,y,1,c);
		}
		query(1,N,1);
		for(int i=1;i<N;i++)
			if(color[i - 1] != color[i])
				if(color[i] > -1)
					ans[color[i]]++;
		for(int i=0;i<N;i++)
			if(ans[i])
				printf("%d %d\n",i,ans[i]);
		puts("");
	}
	return 0;
}