我们发现这是一道模板STL题。
边听歌边debug效率好低。
#include<bits/stdc++.h> #define N 570007 using namespace std; multiset<int> S1,S2; char ch[18]; #define INF ((int)1e9) #define sight(c) (‘0‘<=c&&c<=‘9‘) inline void read(int &x){ static char c;static int b; for(c=getchar(),b=1;!sight(c);c=getchar()) if (c==‘-‘) b=-1; for(x=0;sight(c);c=getchar())x=x*10+c-48; x*=b; } void write(int x){if (x<10){ putchar(48+x);return;} write(x/10),putchar(48+x%10);} inline void writeln(int x){if (x<0) putchar(‘-‘),x*=-1; write(x); putchar(‘\n‘);} int a[N],ed[N],ans,n,m,id,x; multiset<int>::iterator it,itt; void ins(int x){ if (!ans) return; it=S1.upper_bound(x); if (it!=S1.end()) ans=min(ans,abs(x-*it)); if (it!=S1.begin()) ans=min(ans,abs(*(--it)-x)); S1.insert(x); } int main () { // freopen("a.in","r",stdin); // freopen("a.out","w",stdout); read(n); read(m); ans=1e9; S1.insert(-INF),S1.insert(INF); read(a[1]),S1.insert(a[1]);ed[1]=a[1]; for (int i=2;i<=n;i++) read(a[i]),ins(a[i]),ed[i]=a[i],S2.insert(abs(a[i]-a[i-1])); while (m--) { scanf("%s",ch); switch (ch[4]) { case ‘R‘: read(id);read(x);ins(x); if (id!=n) { S2.erase(S2.find(abs(ed[id]-a[id+1]))); S2.insert(abs(x-a[id+1]));} S2.insert(abs(x-ed[id])); ed[id]=x; break; case ‘G‘: writeln(*S2.begin()); break; case ‘S‘: writeln(ans); break; } } }