Initial commit

14lanqiao/test5-1.cpp

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