[POJ 3171] Cleaning Shifts

Description

给出n个小区间[l,r],更新这段区间的代价为c，求覆盖一段区间[m,e]的最小值。

Solution

线段树优化 dp。

线段树维护区间内最小值，即从 m-1 时刻到 i 时刻要用的最少奶牛。

然后右端点排序，每次查询更新即可。

复杂度 O(挺大)

Code

 

// By YoungNeal
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 10005
#define M 86650
#define int long long
using namespace std;

int f[M];
int n,L,R,tot=1;

struct NNode{
    int l,r,val;
    friend bool operator<(NNode a,NNode b){
        return a.r<b.r;
    }
}cow[N];

struct Node{
    int l,r;
    int minn;
    int lch,rch;
}node[M<<2];

void build(int l,int r,int cur){
    if(l==r){
        node[cur].l=node[cur].r=l;
        node[cur].lch=node[cur].rch=-1;
        node[cur].minn=f[l];
        return;
    }
    node[cur].lch=++tot;
    node[cur].rch=++tot;
    int mid=l+r>>1;
    build(l,mid,node[cur].lch);
    build(mid+1,r,node[cur].rch);
    node[cur].l=node[node[cur].lch].l;
    node[cur].r=node[node[cur].rch].r;
    node[cur].minn=min(node[node[cur].lch].minn,node[node[cur].rch].minn);
}

int query(int l,int r,int cur){
    if(l<=node[cur].l&&node[cur].r<=r) return node[cur].minn;
    int mid=node[cur].l+node[cur].r>>1;
    int minn=0x3f3f3f3f;
    if(l<=mid) minn=min(minn,query(l,r,node[cur].lch));
    if(mid<r) minn=min(minn,query(l,r,node[cur].rch));
    return minn;
}

void update(int l,int r,int c,int cur){
    if(l<=node[cur].l&&node[cur].r<=r){
        node[cur].minn=c;
        return;
    }
    int mid=node[cur].l+node[cur].r>>1;
    if(l<=mid) update(l,r,c,node[cur].lch);
    if(mid<r) update(l,r,c,node[cur].rch);
    node[cur].minn=min(node[cur].minn,min(node[node[cur].lch].minn,node[node[cur].rch].minn));
}

signed main(){
    scanf("%lld%lld%lld",&n,&L,&R),L++,R++;
    for(int i=1;i<=n;i++) scanf("%lld%lld%lld",&cow[i].l,&cow[i].r,&cow[i].val),cow[i].l++,cow[i].r++;
    sort(cow+1,cow+1+n);
    memset(f,0x3f,sizeof f);
    f[L-1]=0;
    build(L-1,R,1);
    for(int i=1;i<=n;i++){
        int minn=query(cow[i].l-1,cow[i].r-1,1);
        if(minn+cow[i].val<f[cow[i].r]){
            f[cow[i].r]=minn+cow[i].val;
            update(cow[i].r,cow[i].r,f[cow[i].r],1);
        }
    }
    printf(f[R]>=0x3f3f3f3f?"-1":"%lld\n",f[R]);
    return 0;
}