LA 4256 DP Salesmen

d(i, j)表示使前i个数满足要求，而且第i个数值为j的最小改动次数。

d(i, j) = min{ d(i-1, k) | k == j | G[j][k] }

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 200 + 10;

int n, m, k;
int a[maxn];
int d[maxn][maxn];
bool G[maxn][maxn];

int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        memset(G, false, sizeof(G));
        while(m--)
        {
            int u, v; scanf("%d%d", &u, &v);
            G[u][v] = G[v][u] = true;
        }
        scanf("%d", &k);
        for(int i = 1; i <= k; i++) scanf("%d", a + i);

        memset(d, 0x3f, sizeof(d));
        for(int i = 1; i <= k; i++) d[0][i] = 0;
        for(int i = 1; i <= k; i++)
            for(int j = 1; j <= n; j++)
                for(int t = 1; t <= n; t++) if(G[j][t] || j == t)
                    d[i][j] = min(d[i][j], d[i-1][t] + (j == a[i] ? 0 : 1));

        int ans = k;
        for(int i = 1; i <= n; i++) ans = min(ans, d[k][i]);
        printf("%d
", ans);
    }

    return 0;
}