HDU 1698 Just a Hook

最新推荐文章于 2022-12-12 21:12:45 发布

Wall_F

最新推荐文章于 2022-12-12 21:12:45 发布

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本文详细介绍了线段树成段更新的操作方法，包括更新算法的实现过程及其实例应用，旨在帮助读者理解并掌握这一重要的数据结构操作。

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线段树成段更新。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

const int maxn = 1000010;
const int INF = 0x3f3f3f3f;

int n, m;
int p, v;
int y1, y2;

int sumv[maxn<<2], minv[maxn<<2], maxv[maxn<<2];
int setv[maxn<<2];

void pushup(int o)
{
	sumv[o] = sumv[o*2] + sumv[o*2+1];
}

void build(int o, int L, int R)
{
	int lc = o*2, rc = o*2+1;
	setv[o] = -1;
	sumv[o] = 1;
	if(L == R) return ;
	int M = L + (R-L)/2;
	build(lc, L, M);
	build(rc, M+1, R);
	pushup(o);
}

void maintain(int o, int L, int R)
{
	int lc = o*2, rc = o*2+1;
	if(R > L)
	{
		sumv[o] = sumv[lc] + sumv[rc];
		minv[o] = min(minv[lc], minv[rc]);
		maxv[o] = max(maxv[lc], maxv[rc]);
	}
	if(setv[o] >= 0) { minv[o] = maxv[o] = setv[o]; sumv[o] = setv[o] * (R-L+1);}
}

void pushdown(int o)
{
	int lc = 2*o, rc = 2*o+1;
	if(setv[o] >= 0)
	{
		setv[lc] = setv[rc] = setv[o];
		setv[o] = -1;
	}
}

void update(int o, int L, int R)
{
	int lc = 2*o, rc = 2*o+1;
	if(y1 <= L && y2 >= R)
	{
		setv[o] = v;
	}
	else
	{
		pushdown(o);
		int M = L + (R-L)/2;
		if(y1 <= M) update(lc, L, M); else maintain(lc, L, M);
		if(y2 > M) update(rc, M+1, R); else maintain(rc, M+1, R);
	}
	maintain(o, L, R);
}

int _sum, _min, _max;

void query(int o, int L, int R)
{
	int lc = o*2, rc = o*2+1;
	if(setv[o] >= 0)
	{
		_sum += setv[o] * (min(R, y2) - max(L, y1) + 1);
		_min = min(_min, setv[o]);
		_max = max(_max, setv[o]);
	}
	else if(y1 <= L && y2 >= R)
	{
		_sum += sumv[o];
		_min = min(_min, minv[o]);
		_max = max(_max, maxv[o]);
	}
	else
	{
		int M = L + (R-L)/2;
		if(y1 <= M) query(lc, L, M);
		if(y2 > M) query(rc, M+1, R);
	}
}

void read_case()
{
	scanf("%d", &n);
	scanf("%d", &m);
	build(1, 1, n);
	while(m--)
	{
		scanf("%d%d%d", &y1, &y2, &v);
		update(1, 1, n);
	}
}

void solve()
{
	read_case();
	y1 = 1, y2 = n;
	_min = INF, _max = -INF, _sum = 0;
	query(1, 1, n);
	printf("The total value of the hook is %d.\n", _sum);
}

int main()
{
	int T, times = 0;
	scanf("%d", &T);
	while(T--)
	{
		printf("Case %d: ", ++times);
		solve();
	}
	return 0;
}