Gym 103306J John in the Amusement Park(dp)
作者:互联网
题目链接
题意:
有\(n\)个任务,总时间为\(T\),从\(0\)开始。每个任务有\(m\)个开始时间\(T_i\),有一个高兴值\(h\),持续时间\(t_i\)(每个活动可重复进行,任务活动时间不重叠),且最后一个任务在\(T\)时间内开始,不必\(T\)时间内结束。
求\(T\)时间内最大高兴值。
思路:
动态规划
\(dp[i]:\)某活动在时刻\(i\)结束,时间\(i\)内最大高兴值,否则为\(0\)。
对于任务\(K\),开始时间为\(s_k\),结束时间为\(e_k\),高兴值为\(w\),则\(dp[e_k] = max(max(dp[m]+w),dp[e_k]),m<=s_k\)。
\(ans=max(dp[i]),i<=T\)
code:
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <deque>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#include <unordered_map>
#define fi first
#define se second
#define pb push_back
#define endl "\n"
#define debug(x) cout << #x << ":" << x << endl;
#define bug cout << "********" << endl;
#define all(x) x.begin(), x.end()
#define lowbit(x) x & -x
#define fin(x) freopen(x, "r", stdin)
#define fout(x) freopen(x, "w", stdout)
#define ull unsigned long long
#define ll long long
const double eps = 1e-15;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
using namespace std;
struct node{
int s, e, h;
bool operator<(const node &a)const{
return s < a.s;
}
}s[maxn];
int dp[maxn], n, t;
int main(){
scanf("%d%d" ,&n, &t);
int tot = 0;
for(int i = 1; i <= n; i ++){
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
for(int j = 1; j <= c; j ++){
scanf("%d", &s[++ tot].s);
s[tot].e = s[tot].s + b, s[tot].h = a;
}
}
sort(s + 1, s + tot + 1);
int maxx = 0, p = 1;
for(int i = 0; i <= t; i ++){
maxx = max(maxx, dp[i]);
while(p <= tot && s[p].s == i){
if(s[p].e <= t)dp[s[p].e] = max(dp[s[p].e], maxx + s[p].h);
else dp[t] = max(dp[t], maxx + s[p].h);
p ++;
}
}
printf("%d\n", max(maxx, dp[t]));
return 0;
}
标签:103306J,max,Gym,Park,任务,时间,include,dp,define 来源: https://www.cnblogs.com/lniiwuw/p/15363160.html