f[i] represents the longest idle time of i~n;
If there is no current task, take a rest for one second (f[i]=f[i+1]+1);
otherwise f[i]=max(f[i],f[i+current working time]);
Using the structure to record, we open an array for each moment to store all the tasks starting at this moment, and the subscript 0 represents the number of elements in the array, which is easy to traverse:
#include<iostream>//简短的30行代码 #include<cstdio> #include<algorithm> #include<cstdio> using namespace std; int n,k,f[10010]; struct cym{ int num[10010]; }a[10010]; int main() { scanf("%d%d",&n,&k); for(int i=1;i<=k;i++) { int x,y; scanf("%d%d",&x,&y); a[x].num[ ++a[x].num[ 0 ]]= y; } for(int i=n;i>=0;i--) { if(!a[i].num[0]) { f[i]=f[i+1]+1; continue; } for(int j=1;j<=a[i].num[0];j++) f[i]=max(f[i],f[i+a[i].num[j]]); } printf("%d",f[1]); }