Linear Kingdom Races CodeForces - 115E (线段树优化dp)

大意: n条赛道, 初始全坏, 修复第$i$条花费$a_i$, m场比赛, 第$i$场比赛需要占用$[l_i,r_i]$的所有赛道, 收益为$w_i$, 求一个比赛方案使得收益最大.

设$dp[i]$为只考虑前$i$条赛道的最大收益, $calc(i,j)$为占用区间$[i,j]$的赛道的比赛收益和, $s$为$a$的前缀和, 有

$$dp[i]=\max\limits_{1\le j < i}(dp[j]+calc(j+1,i)+s[j])-s[i]$$

$calc$的贡献用线段树更新即可, 水题一道.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) {REP(i,1,n) cout<<a[i]<<' ';hr;}
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head

#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, m, a[N];
ll s[N], dp[N], f[N];
struct _ {
	int l,r,w;
	bool operator < (const _ & rhs) const {
		return r<rhs.r;
	}
} q[N];
ll v[N<<2], tag[N<<2];
void pd(int o) {
	if (tag[o]) {
		v[lc]+=tag[o],tag[lc]+=tag[o];
		v[rc]+=tag[o],tag[rc]+=tag[o];
		tag[o]=0;
	}
}
void add(int o, int l, int r, int ql, int qr, ll w) {
	if (ql<=l&&r<=qr) return v[o]+=w,tag[o]+=w,void();
	pd(o);
	if (mid>=ql) add(ls,ql,qr,w);
	if (mid<qr) add(rs,ql,qr,w);
	v[o]=max(v[lc],v[rc]);
}
ll qry(int o, int l, int r, int ql, int qr) {
	if (ql<=l&&r<=qr) return v[o];
	pd(o);
	ll ans = 0;
	if (mid>=ql) ans=max(ans,qry(ls,ql,qr));
	if (mid<qr) ans=max(ans,qry(rs,ql,qr));
	return ans;
}

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,n) scanf("%d", a+i),s[i]=s[i-1]+a[i];
	REP(i,1,m) scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].w);
	sort(q+1,q+1+m);
	int now = 1;
	REP(i,1,n) {
		while (now<=m&&q[now].r==i) {
			add(1,0,n,0,q[now].l-1,q[now].w);
			++now;
		}
		dp[i] = max(dp[i-1], qry(1,0,n,0,i-1)-s[i]);
		add(1,0,n,i,i,dp[i]+s[i]);
	}
	printf("%lld\n", dp[n]);
}