递推DP UVA 1424 Salesmen

题目传送门

 /*
    题意：给定包含n个点的无向图和一个长度为L的序列，修改尽量少的点使得相邻的数字相同或连通
    DP：状态转移方程：dp[i][j] = min (dp[i][j], dp[i-1][k] + (j != a[i]));
         dp[i][j]表示前i个数字以j结尾的最小花费。我自己写了很长时间，很复杂，状态转移的不好。
         应该能知道前一个状态的所有情况，每一维数组记录的就是一个状态
 */
 /************************************************
 Author        :Running_Time
 Created Time  :2015-8-5 9:03:34
 File Name     :UVA_1424.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e2 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 bool con[MAXN][MAXN];
 int a[MAXN*];
 int dp[MAXN*][MAXN];
 int n1, m, n;

 int work(void)  {
     memset (dp, INF, sizeof (dp));
     for (int i=; i<=n1; ++i)   {
         dp[][i] = (i != a[]);
     }
     for (int i=; i<=n; ++i)    {
         for (int j=; j<=n1; ++j)   {
             for (int k=; k<=n1; ++k)   {
                 if (con[j][k])  {
                     dp[i][j] = min (dp[i][j], dp[i-][k] + (j != a[i]));
                 }
             }
         }
     }
     int res = INF;
     for (int i=; i<=n1; ++i)    {
         res = min (res, dp[n][i]);
     }

     return res;
 }

 int main(void)    {     //UVA 1424 Salesmen
     int T;  scanf ("%d", &T);
     while (T--) {
         scanf ("%d%d", &n1, &m);
         memset (con, false, sizeof (con));
         for (int i=; i<=n1; ++i)   con[i][i] = true;
         for (int i=; i<=m; ++i)    {
             int u, v;   scanf ("%d%d", &u, &v);
             con[u][v] = con[v][u] = true;
         }
         scanf ("%d", &n);
         for (int i=; i<=n; ++i)    {
             scanf ("%d", &a[i]);
         }

         printf ("%d\n", work ());
     }

     return ;
 }

 /************************************************
 Author        :Running_Time
 Created Time  :2015-8-5 9:03:34
 File Name     :UVA_1424.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 1e2 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 bool con[MAXN][MAXN];
 vector<int> G[MAXN];
 vector<int> pre;
 int a[MAXN*];
 int dp[MAXN][][MAXN];
 int n1, m, n;

 int work(void)  {
     memset (dp, INF, sizeof (dp));    pre.clear ();
     dp[][][a[]] = ;
     for (int i=; i<=n1; ++i)   {
         if (i == a[])  continue;
         pre.push_back (i);
         dp[][][i] = ;
     }
     for (int i=; i<=n; ++i)    {
         if (con[a[i-]][a[i]])  {
             dp[i][][a[i]] = dp[i-][][a[i-]];
             pre.clear ();   continue;
         }
         vector<int> tmp;
         for (int j=; j<pre.size (); ++j)   {
             int u = pre[j];
             for (int k=; k<G[u].size (); ++k) {
                 int v = G[u][k];
                 if (con[u][v])  {
                     if (v == a[i])  dp[i][][v] = min (dp[i][][v], dp[i-][][u]);
                     else    dp[i][][v] = min (dp[i][][v], dp[i-][][u] + );
                     tmp.push_back (v);
                 }
             }
         }
         pre.clear ();
         for (int j=; j<tmp.size (); ++j)   pre.push_back (tmp[j]);
     }

     int res = INF;
     res = min (res, dp[n][][a[n]]);
     for (int i=; i<pre.size (); ++i)   {
         int v = pre[i];
         res = min (res, dp[n][][v]);
     }

     //debug
     for (int i=; i<=n; ++i)    {
         printf ("%d ", dp[i][][a[i]]);
     }
     puts ("");

     return res;
 }

 int main(void)    {
     int T;  scanf ("%d", &T);
     while (T--) {
         scanf ("%d%d", &n1, &m);
         memset (con, false, sizeof (con));
         for (int i=; i<=n1; ++i)   G[i].clear ();
         for (int i=; i<=n1; ++i)   con[i][i] = true;
         for (int i=; i<=m; ++i)    {
             int u, v;   scanf ("%d%d", &u, &v);
             con[u][v] = con[v][u] = true;   G[u].push_back (v); G[v].push_back (u);
         }
         scanf ("%d", &n);
         for (int i=; i<=n; ++i)    {
             scanf ("%d", &a[i]);
         }

         printf ("%d\n", work ());
     }

     return ;
 }

undone(感受一下就行了)