把坐标离散化之后就是很普通的尺取法啦。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-9 # define MOD 100000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=1000005; //Code begin... struct Node{int c; LL pos;}node[N]; int vis[65], num; bool comp(Node a, Node b){return a.pos<b.pos;} int main () { LL ans=(LL)1<<60; int n, k, t; n=Scan(); k=Scan(); FOR(i,1,k) { t=Scan(); while (t--) node[++num].pos=Scan(), node[num].c=i; } sort(node+1,node+n+1,comp); num=0; int l=1, r=0; while (1) { while (num<k) { ++r; if (r>n) break; if (vis[node[r].c]==0) ++num; ++vis[node[r].c]; } if (r>n) break; ans=min(ans,node[r].pos-node[l].pos); --vis[node[l].c]; if (vis[node[l].c]==0) --num; ++l; } printf("%lld ",ans); return 0; }