CF-45D. Event Dates(贪心)
On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn't remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.
The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and rieach (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.
Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.
3 1 2 2 3 3 4
1 2 3
2 1 3 1 3
1 2
思路:每次选择左边界最小,次之右边界最小,选择此区间的左边界值l作为此区间代表,并删去本区间,然后提升剩余区间的左边界使之大于那个区间的代表l,重复此步骤即可
#include<iostream> #include<cstring> using namespace std; const int mm=110; class node { public:int l,r; }f[mm]; bool vis[mm]; int ans[mm]; bool ok(int a,int b) { if(f[a].l<f[b].l)return 1; else if(f[a].l==f[b].l&&f[a].r<f[b].r)return 1; return 0; } int main() { int n; while(cin>>n) { for(int i=0;i<n;i++) { cin>>f[i].l>>f[i].r;vis[i]=0; } int k; for(int i=0;i<n;i++) { k=-1; for(int j=0;j<n;j++) if(!vis[j]&&(k==-1||ok(j,k)))///选择区间左界最小 k=j; int w=ans[k]=f[k].l; ++w;vis[k]=1; for(int j=0;j<n;j++)///提升剩余区间的左界 if(w>f[j].l) f[j].l=w; } for(int i=0;i<n;i++) cout<<ans[i]<<" ";cout<<"\n"; } }