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] }

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5
 6 const int maxn = 200 + 10;
 7
 8 int n, m, k;
 9 int a[maxn];
10 int d[maxn][maxn];
11 bool G[maxn][maxn];
12
13 int main()
14 {
15     int T; scanf("%d", &T);
16     while(T--)
17     {
18         scanf("%d%d", &n, &m);
19         memset(G, false, sizeof(G));
20         while(m--)
21         {
22             int u, v; scanf("%d%d", &u, &v);
23             G[u][v] = G[v][u] = true;
24         }
25         scanf("%d", &k);
26         for(int i = 1; i <= k; i++) scanf("%d", a + i);
27
28         memset(d, 0x3f, sizeof(d));
29         for(int i = 1; i <= k; i++) d[0][i] = 0;
30         for(int i = 1; i <= k; i++)
31             for(int j = 1; j <= n; j++)
32                 for(int t = 1; t <= n; t++) if(G[j][t] || j == t)
33                     d[i][j] = min(d[i][j], d[i-1][t] + (j == a[i] ? 0 : 1));
34
35         int ans = k;
36         for(int i = 1; i <= n; i++) ans = min(ans, d[k][i]);
37         printf("%d\n", ans);
38     }
39
40     return 0;
41 }

代码君