比较简单的dp。
转移方程是:
dp[i] = Max(dp[i],dp[j]+a[i].ef) //0<=j< i,条件是a[i].st>=a[j].ed
#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
#include<set>
#include<bitset>
//#define ONLINE_JUDGE
#define eps 1e-8
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
#define ll __int64
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
//#pragma comment(linker,"/STACK:1024000000,1024000000")
int n,m;
#define M 110
#define N 1000010
#define Mod 1000000000
#define p(x,y) make_pair(x,y)
const int MAX_len=550;
struct Node{
ll st,ed,ef;
}a[1010];
bool cmp(Node x,Node y){
return x.st<y.st;
}
ll dp[1010];
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
#endif
int r;
while(scanf("%d%d%d",&n,&m,&r)!=EOF){
for(int i=0;i<m;i++){
scanf("%I64d%I64d%I64d",&a[i].st,&a[i].ed,&a[i].ef);
a[i].ed += r; //需要休息r分钟
}
sort(a,a+m,cmp);
for(int i=0;i<m;i++){
dp[i] = a[i].ef; //初始化
for(int j=0;j<i;j++){
if(a[i].st>=a[j].ed){ //当a[i].st>=a[j].ed时,可以取dp[i]和dp[j]+a[i].ef中大的那个
dp[i] = Max(dp[i],dp[j]+a[i].ef);
}
}
}
ll ans=-1;
for(int i=0;i<m;i++)
ans = Max(ans,dp[i]);
printf("%I64d\n",ans);
}
return 0;
}