Codeforces Round #364 (Div.2) C:They Are Everywhere(双指针/尺取法)
题目链接:
题意:
给出一个长度为n的字符串,要我们找出最小的子字符串包含所有的不同字符。
分析:
1.尺取法,不懂得可以看今天开始学算法系列-尺取法
2.二分法,具体见代码
PS:因为讲得太少所以附上A和B的题解:
A:结构体存储,将所有数sort一下,一头一尾输出即可
B:开辆个数组记录当前行或列是否有车,row[],col[],设两个变量a,b记录行列,放入一个车时,如果已标记则不用减去,否则减去(a--或b--),输出a*b即可。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
![](https://images.cnblogs.com/OutliningIndicators/ExpandedBlockStart.gif)
//尺取法 #include <algorithm> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <ctime> #include <deque> #include <functional> #include <iostream> #include <list> #include <map> #include <queue> #include <set> #include <stack> #include <string> #include <vector> using namespace std; char inp[100010]; int ch[300]; int main() { int n,s,e,i,poke = 0,curpoke = 0,res = n; scanf("%d",&n);getchar();scanf("%s",inp); for(i = 0; i < n; i++) ch[inp[i]]++; for(i = 0; i < 256; i++)if(ch[i])poke++; for(i = 0; i < 256; i++)ch[i] = 0; s=e=0; while(s <= e && e <= n) { if(curpoke < poke) { if(ch[inp[e]] == 0) curpoke++; ch[inp[e]]++; e++; } else { if(ch[inp[s]] == 1) curpoke--; ch[inp[s]]--; s++; } if(poke == curpoke) res = min(res, e - s); } printf("%d",res); }
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
![](https://images.cnblogs.com/OutliningIndicators/ExpandedBlockStart.gif)
//二分求解 #include<iostream> #include<cstring> #include<cstdio> #define ll __int64 #define PI acos(-1.0) #define mod 1000000007 using namespace std; int n; map<char,int> mp; map<char,int> gg; char a[100005]; int zong; bool check(int xx) { int zha=0; gg.clear(); for(int i=1;i<=xx;i++) { if(gg[a[i]]==0) zha++; gg[a[i]]++; } if(zha>=zong) return true; for(int i=xx+1;i<=n;i++) { if(gg[a[i]]==0) zha++; gg[a[i]]++; gg[a[i-xx]]--; if(gg[a[i-xx]]==0) zha--; if(zha>=zong) return true; } return false; } int main() { zong=0; scanf("%d",&n); getchar(); mp.clear(); for(int i=1;i<=n;i++) { scanf("%c",&a[i]); if(mp[a[i]]==0) { mp[a[i]]=1; zong++; } } int l=zong,r=n,mid=0; while(l<r) { mid=(l+r)/2; if(check(mid)) r=mid; else l=mid+1; } printf("%d\n",l); return 0; }
一直地一直地往前走