[51NOD1127]最短的包含字符串（尺取法）

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1127

思路：尺取法，一开始我考虑更新右指针，直到遇到一个和l指针指向的字符相同的时候为止，发现这样做ac不了。于是换了一个思路。

一直更新r指针，直到所有字符都出现了一遍后，更新答案和左指针，导致有一个缺口，这时候再更新r指针。

     /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%lld", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 int ascii[];
 char s[maxn];
 int n;

 int main() {
     // FRead();
     while(~Rs(s)) {
         Cls(ascii);
         n = strlen(s);
         if(n < ) {
             printf("No Solution\n");
             continue;
         }
         int ret = 0x7f7f7f;
         int cnt = ;
         int l = , r = ;
         ascii[s[l]]++; cnt = ;
         while(r < n) {
             if(ascii[s[r]] == ) cnt++;
             ascii[s[r]]++;
             while(cnt == ) {
                 ret = min(ret, r-l+);
                 ascii[s[l]]--;
                 if(ascii[s[l]] == ) cnt--;
                 l++;
             }
             r++;
         }
         if(ret == 0x7f7f7f) printf("No Solution\n");
         else printf("%d\n", ret);
     }
     RT ;
 }