33 lines
822 B
C++
33 lines
822 B
C++
#include<bits/stdc++.h>
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using namespace std;
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const int N = 1e5 + 10;
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int ft[N], bk[N], dp[N];
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/*
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ft[i]第i个元素的首数字、bk［i］第i个元素尾数字
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类LIS问题的DP解法-时间复杂度0(n^2)
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状态 dp[i] 以第主个元素作为结尾的最长接龙数列的长度
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状态转移方程 dp[i] = max(dp[i], dp[j]+1);
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*/
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int main(){
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int n; cin >> n;
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string s;
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for(int i = 1; i<=n; i++){
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cin >> s;
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ft[i] = s.front() - '0', bk[i] = s.back() - '0', dp[i] = 1;
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}
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int mx = 1;
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for(int i = 2; i <= n; i++){
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for(int j = 1; j < i; j++){
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if(bk[j] == ft[i]){
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dp[i] = max(dp[i], dp[j] + 1);
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}
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mx = max(mx, dp[i]);
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}
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}
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cout << n - mx << endl;
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return 0;
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}
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/* test samples 易超时
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5
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11 121 22 12 2023
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*/ |