topic
train of thought
The knowledge point of the test is the line segment tree + dp .
We can follow the dp 4-step method to deduce step by step,
1.dp definition
dp[i] represents the minimum cost for the [1,i] interval to be covered by a given line segment
2. State transition equation
According to the definition of dp, it can be enumerated to the xth line , the interval is [Lx, Rx] , and the cost is Vx
dp[Rx]=min(dp[i](i = (Lx - 1) ~ Rx))+Vx
Because when you want to cover the interval of [1, Rx], you must add the xth line segment on the basis of covering [1, Lx - 1]/[1, Lx].../[1, Rx] .
And the minimum cost of covering [1,Lx - 1]/[1,Lx].../[1,Rx] is (dp[i](i = (Lx - 1) ~ Rx) , plus the xth item The cost of a line segment is Vx .
Draw a picture to understand:
这个时候会有个问题:在这里我们要求min(dp[i](i = (Lx - 1) ~ Rx)，可是如果暴力枚举找min的话时间复杂度O(n*m)会超时,那该怎么办呢?我们可以用线段树(详见详解线段树)来实现查找区间最小值和单点更新(dp在变)的效果。
3.初始化
一个是dp一开始要是无穷大,一个是minv(线段树的那个)也要初始化为无穷大,另外储存线段的数组也要以r从大到小排序(这样才能保证后续枚举线段时dp[i](i = (Lx - 1) ~ Rx是最优的)
4.遍历顺序
显然就是以下标顺序枚举所有线段。
代码
#include <bits/stdc++.h>
#define int long long
#define maxn 300000
using namespace std;
int n,m,dp[5000000],minv[5000000];
struct shui
{
int l,r,v;
} a[5000000];
bool cmp(shui x,shui y)
{
return x.r < y.r;
}
int find(int id,int l,int r,int x,int y)//查找[x,y]区间内的最小值
{
if(x <= l && r <= y) return minv[id];
int mid = (l + r) / 2,ans = INT_MAX;
if(x <= mid) ans = min(ans,find(id * 2,l,mid,x,y));
if(y > mid) ans = min(ans,find(id * 2 + 1,mid + 1,r,x,y));
return ans;
}
void gexi(int id,int l,int r,int x,int v)//将编号为x的节点的值更改为v
{
if(l == r)
{
minv[id] = v;
return ;
}
int mid = (l + r) / 2;
if(x <= mid) gexi(id * 2,l,mid,x,v);
else gexi(id * 2 + 1,mid + 1,r,x,v);
minv[id] = min(minv[id * 2],minv[id * 2 + 1]);
}
signed main()
{
scanf("%lld%lld",&n,&m);
for(int i = 1; i <= n; i++) scanf("%lld%lld%lld",&a[i].l,&a[i].r,&a[i].v);
memset(dp,0x3f,sizeof(dp));
memset(minv,0x3f,sizeof(minv));//minv对应的是dp
sort(a + 1,a + 1 + n,cmp);
for(int i = 1; i <= n; i++)
{
if(a[i].l == 1)
{
if(dp[a[i].r] > a[i].v)
{
dp[a[i].r] = a[i].v;
gexi(1,1,maxn,a[i].r,a[i].v);
}
}
else
{
int t = find(1,1,maxn,a[i].l - 1,a[i].r);
if(dp[a[i].r] > (t + a[i].v))
{
dp[a[i].r] = t + a[i].v;
gexi(1,1,maxn,a[i].r,dp[a[i].r]);
}
}
}
printf("%lld",dp[m]);
return 0;
}