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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 
#include 
#include 
#include 
#include 
using namespace std;
const int MAXN = 1005;

int a[MAXN],n,m,num;
int dp[MAXN][MAXN];
vector 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;
}