Luogu P1280 尼可的任务(dp)

 

题意：

时间为n,有k个任务，每个任务有一个开始时间和持续时间，从第一分钟开始，如果有开始的任务就要做，问最大空闲时间

n,k<=1e5

思路：

设 dp[i]为i~n时间中最大空闲时间，vector v[i]保存i开始的任务，倒推即可

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x))

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e3+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int n, k;
int a[maxn];
vector<int>v[maxn];
int dp[maxn];
int main(){
    scanf("%d %d", &n, &k);
    for(int i = 1; i <= k; i++){
        int x, y ;
        scanf("%d %d", &x, &y);
        v[x].pb(y);
    }
    for(int i = n; i >= 1; i --){
        if(v[i].size()==0){
            dp[i] = dp[i+1]+1;
        }
        else{
            for(int j = 0; j < (int)v[i].size(); j++){
                dp[i] = max(dp[i+v[i][j]],dp[i]);
            }
        }
    }
    printf("%d",dp[1]);
    return 0;
}
/**/