Educational Codeforces Round 67 (Rated for Div. 2)B. Letters Shop
题意：找到从头开始最短的串，使得串的字母个数涵盖给出子串所有字母的个数
做法：一开始直接暴力计数加查找，后来TLE了才想到用二分。。。我真的是傻了
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<queue> #include<map> #define ll long long #define pb push_back #define rep(x,a,b) for (int x=a;x<=b;x++) #define repp(x,a,b) for (int x=a;x<b;x++) #define W(x) printf("%d\n",x) #define WW(x) printf("%lld\n",x) #define pi 3.14159265358979323846 #define mem(a,x) memset(a,x,sizeof a) #define lson rt<<1,l,mid #define rson rt<<1|1,mid+1,r using namespace std; const int maxn=2e5+7; const int INF=1e9; const ll INFF=1e18; char s[maxn]; char s1[maxn]; int num[27][maxn];//记录原始串在第i个位置之前各个字母的数量 int num1[27];//记录所求串的字母个数 int main() { int l,m; scanf("%d",&l); scanf("%s",s); rep(i,0,l-1)//打表 { repp(j,0,26) if (i!=0)num[j][i]=num[j][i-1]; num[s[i]-'a'][i]++; } scanf("%d",&m); while(m--) { scanf("%s",s1); rep(i,0,26)//记得清空 num1[i]=0; int l1=strlen(s1); repp(i,0,l1) num1[s1[i]-'a']++; int ans=0; repp(i,0,26) { int left=0,r=l-1,mid; if (num1[i]==0)continue; while(left<r) { mid=(left+r)>>1; if (num[i][mid]>=num1[i])r=mid; else left=mid+1; } ans=max(ans,left+1); } W(ans); } return 0; } 来源：51CTO
作者：w_udixixi
链接：https://blog.csdn.net/w_udixixi/article/details/101223043