Lunar New Year and Red Envelopes CodeForces - 1106E (dp)
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大意: 总共$n$的时间, $k$个红包, 红包$i$只能在时间$[s_i,t_i]$范围内拿, 并且拿完后时间跳到$d_i+1$,Bob采用贪心策略,每个时间点若有红包能取则取钱数$w_i$最大的, $w_i$相同则取$d_i$最大的, Alice有$m$次机会让Bob跳过一个时间, 求Alice如何操作能使Bob得到钱数最少.
比较简单的DP, 记$dp[i][j]$为$i$时刻还能干扰$j$次的最小收益
$dp[i][j]=min(dp[d_i][j]+w_i,dp[i+1][j-1])$, $(w_i,d_i)$为$i$时刻的贪心策略
用一个堆维护一个最优的(w,d)来转移即可
#include <iostream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #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 w first #define d second #define io std::ios::sync_with_stdio(false) #define endl ‘ ‘ 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 const int N = 1e5+10; int n, m, k, t; struct _ { int s,t,d,w; bool operator < (const _ &rhs) const { return pii(w,d) < pii(rhs.w,rhs.d); } } a[N]; ll dp[N][202]; priority_queue<_> q; int main() { scanf("%d%d%d", &n, &m, &k); REP(i,1,k) { scanf("%d%d%d%d",&a[i].s,&a[i].t,&a[i].d,&a[i].w); } sort(a+1,a+1+k,[](_ a,_ b){return a.t<b.t;}); memset(dp,0x3f,sizeof dp); REP(i,0,m) dp[n+1][i]=0; int now = k; PER(i,1,n) { while (now>0&&a[now].t>=i) q.push(a[now--]); while (q.size()&&q.top().s>i) q.pop(); if (q.empty()) { REP(j,0,m) dp[i][j]=dp[i+1][j]; continue; } int d=q.top().d,w=q.top().w; dp[i][0] = dp[d+1][0]+w; REP(j,1,m) dp[i][j]=min(dp[d+1][j]+w,dp[i+1][j-1]); } printf("%lld ", dp[1][m]); }
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