题意：给定一个包含n个点的无向连通图和一个长为L的序列，你的任务是修改尽量少的点
，使得序列中的任意两个相邻的数或者相同，或者相邻
思路：不算太难想到的DP。dp[i][j]表示第i个是j 的最少的修改数，然后每次都枚举它的相邻点，所以dp[i][j]=min(dp[i][j],dp[i-1][k])
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int MAXN = 1005;
int a[MAXN],n,m,num;
int dp[MAXN][MAXN];
vector<int> arr[MAXN];
int main(){
int t;
scanf("%d",&t);
while (t--){
scanf("%d%d",&n,&m);
for (int i = 1; i<= n; i++)
arr[i].clear(),arr[i].push_back(i);
for (int i = 1; i <= m; i++){
int u,v;
scanf("%d%d",&u,&v);
arr[u].push_back(v);
arr[v].push_back(u);
}
scanf("%d",&num);
memset(dp,0x3f3f3f3f,sizeof(dp));
for (int i = 1; i <= num; i++)
scanf("%d",&a[i]);
for (int i = 1; i <= n; i++){
if (i == a[1])
dp[1][i] = 0;
else dp[1][i] = 1;
}
for (int i = 2; i <= num; i++){
for (int j = 1; j <= n; j++){
int len = arr[j].size();
for (int k = 0; k < len; k++){
int temp = arr[j][k];
dp[i][j] = min(dp[i][j],dp[i-1][temp]);
}
if (j != a[i])
dp[i][j]++;
}
}
int Min = 0x3f3f3f3f;
for (int i = 1; i <= n; i++)
Min = min(Min,dp[num][i]);
printf("%d\n",Min);
}
return 0;
}