51nod 1127 最短的包含字符串(尺取法)
http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1127
可以用一个映射,把从头到当前位置的出现了多少个不同的字符计算出来,不断向区间末尾推进,如果包含了A-Z就可以更新最短长度,然后把开头的字符减1,表示向前推进一位.
这样就能在nlogn的时间内出解. 用map或者一个一维数组来映射都可以.
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <vector> 5 #include <cstring> 6 #include <string> 7 #include <algorithm> 8 #include <string> 9 #include <set> 10 #include <functional> 11 #include <numeric> 12 #include <sstream> 13 #include <stack> 14 #include <map> 15 #include <queue> 16 #pragma comment(linker, "/STACK:102400000,102400000") 17 #define CL(arr, val) memset(arr, val, sizeof(arr)) 18 19 #define ll long long 20 #define inf 0x7f7f7f7f 21 #define lc l,m,rt<<1 22 #define rc m + 1,r,rt<<1|1 23 #define pi acos(-1.0) 24 25 #define L(x) (x) << 1 26 #define R(x) (x) << 1 | 1 27 #define MID(l, r) (l + r) >> 1 28 #define Min(x, y) (x) < (y) ? (x) : (y) 29 #define Max(x, y) (x) < (y) ? (y) : (x) 30 #define E(x) (1 << (x)) 31 #define iabs(x) (x) < 0 ? -(x) : (x) 32 #define OUT(x) printf("%I64d\n", x) 33 #define lowbit(x) (x)&(-x) 34 #define Read() freopen("a.txt", "r", stdin) 35 #define Write() freopen("b.txt", "w", stdout); 36 #define maxn 1000000000 37 #define N 510 38 #define mod 1000000000 39 using namespace std; 40 char str[100010]; 41 int main() 42 { 43 //Read(); 44 scanf("%s",str); 45 int l=strlen(str); 46 map<char,int>m; 47 int s=0,t=0,num=0,res=inf; 48 while(1) 49 { 50 while(t<l&&num<26) 51 { 52 if(m[str[t++]]++==0) num++; 53 //printf("%d\n",num); 54 } 55 if(num<26) break; 56 res=min(res,t-s); 57 if(--m[str[s++]]==0) num--; 58 } 59 if(res==inf) printf("No Solution\n"); 60 else printf("%d\n",res); 61 return 0; 62 }
