2019杭电多校一 C. Milk (dp)

大意: $n*m$棋盘, 初始位置$(1,1)$, 横坐标为$\frac{m+1}{2}$时可以向下走, 否则只能左右走, 每走一步花费$1$秒. 有$k$管奶, 第$i$罐位置$(r_i,c_i)$, 要花费$t_i$的时间去喝. 对于所有的$1\le i\le k$, 求出喝完$i$管奶最短用时.

实现略复杂的$dp$题, 按照官方题解写了.

主要思路是对每一行求出向左/向右喝$x$罐奶的最少用时, 然后$dp$合并答案.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#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 x first
#define y second
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const ll INF = 1e15;
const int N = 1e4+10;
int n, m, k, b[N];
struct {int x,y,t;} a[N];
ll ans[N];
vector<pii> v[N];
vector<ll> solve(vector<pii> a, int s, int x) {
	a.insert(a.begin(),pii(s,0));
	vector<ll> ret(a.size(),INF), t(ret);
	ret[0] = x==-1?0:abs(x-s);
	t[0] = 0;
	for (int i=1; i<a.size(); ++i) {
		int len = abs(a[i].x-a[i-1].x);
		REP(j,0,i-1) t[j]+=len;
		PER(j,1,i) t[j]=min(t[j],t[j-1]+a[i].y);
		REP(j,1,i) ret[j]=min(ret[j],t[j]+(x==-1?0:abs(x-a[i].x)));
	}
	return ret;
}
vector<ll> Merge(const vector<ll> &L, const vector<ll> &R) {
	vector<ll> ret(L.size()+R.size()-1,INF);
	for (int i=0;i<L.size();++i) {
		for (int j=0;j<R.size();++j) {
			ret[i+j] = min(ret[i+j], L[i]+R[j]);
		}
	}
	return ret;
}

void work() {
	scanf("%d%d%d", &n, &m, &k);
	*b = 0;
	REP(i,1,k) {
		scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].t);
		b[++*b] = a[i].x;
	}
	b[++*b] = 1;
	sort(b+1,b+1+*b),*b=unique(b+1,b+1+*b)-b-1;
	REP(i,1,*b) v[i].clear();
	REP(i,1,k) {
		a[i].x=lower_bound(b+1,b+1+*b,a[i].x)-b;
		v[a[i].x].pb(pii(a[i].y,a[i].t));
		ans[i] = INF;
	}
	REP(i,1,*b) sort(begin(v[i]),end(v[i]));
	auto g = solve(v[1],1,-1);
	for (int i=0;i<g.size();++i) {
		ans[i] = min(ans[i], g[i]);
	}
	int cur = 0;
	vector<ll> f[2];
	f[0] = solve(v[1],1,(m+1)/2);
	REP(i,2,*b) {
		cur ^= 1;
		auto p = lower_bound(begin(v[i]),end(v[i]),pii((m+1)/2,0));
		vector<pii> L(begin(v[i]),p), R(p,end(v[i]));
		reverse(begin(L),end(L));
		auto f0 = solve(L,(m+1)/2,(m+1)/2), f1 = solve(R,(m+1)/2,(m+1)/2);
		auto g0 = solve(L,(m+1)/2,-1), g1 = solve(R,(m+1)/2,-1);
		g0 = Merge(f1,g0), g1 = Merge(f0,g1);
		auto g = g0;
		for (int j=0;j<g.size();++j) {
			g[j] = min(g[j], g1[j]);
		}
		g = Merge(f[!cur],g);
		f[cur] = Merge(f[!cur],Merge(f0,f1));
		for (int j=0;j<g.size();++j) {
			ans[j] = min(ans[j], g[j]+b[i]-b[i-1]);
			f[cur][j] += b[i]-b[i-1];
		}
	}
	REP(i,1,k) {
		if (i==k) printf("%lld\n",ans[i]);
		else printf("%lld ",ans[i]);
	}
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) work();
}

原文地址：https://www.cnblogs.com/uid001/p/11249158.html