递推DP UVA 1424 Salesmen

 /*
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 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
bool con[MAXN][MAXN];
int a[MAXN*2];
int dp[MAXN*2][MAXN];
int n1, m, n;

int work(void)  {
memset (dp, INF, sizeof (dp));
for (int i=1; i<=n1; ++i)   {
dp[1][i] = (i != a[1]);
}
for (int i=2; i<=n; ++i)    {
for (int j=1; j<=n1; ++j)   {
for (int k=1; k<=n1; ++k)   {
if (con[j][k])  {
dp[i][j] = min (dp[i][j], dp[i-1][k] + (j != a[i]));
}
}
}
}
int res = INF;
for (int i=1; i<=n1; ++i)    {
res = min (res, dp
[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=1; i<=n1; ++i)   con[i][i] = true;
for (int i=1; i<=m; ++i)    {
int u, v;   scanf ("%d%d", &u, &v);
con[u][v] = con[v][u] = true;
}
scanf ("%d", &n);
for (int i=1; i<=n; ++i)    {
scanf ("%d", &a[i]);
}

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

return 0;
}

 /************************************************
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 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
bool con[MAXN][MAXN];
vector<int> G[MAXN];
vector<int> pre;
int a[MAXN*2];
int dp[MAXN][2][MAXN];
int n1, m, n;

int work(void)  {
memset (dp, INF, sizeof (dp));    pre.clear ();
dp[1][0][a[1]] = 0;
for (int i=1; i<=n1; ++i)   {
if (i == a[1])  continue;
pre.push_back (i);
dp[1][1][i] = 1;
}
for (int i=2; i<=n; ++i)    {
if (con[a[i-1]][a[i]])  {
dp[i][0][a[i]] = dp[i-1][0][a[i-1]];
pre.clear ();   continue;
}
vector<int> tmp;
for (int j=0; j<pre.size (); ++j)   {
int u = pre[j];
for (int k=0; k<G[u].size (); ++k) {
int v = G[u][k];
if (con[u][v])  {
if (v == a[i])  dp[i][0][v] = min (dp[i][0][v], dp[i-1][1][u]);
else    dp[i][1][v] = min (dp[i][1][v], dp[i-1][1][u] + 1);
tmp.push_back (v);
}
}
}
pre.clear ();
for (int j=0; j<tmp.size (); ++j)   pre.push_back (tmp[j]);
}

int res = INF;
res = min (res, dp
[0][a
]);
for (int i=0; i<pre.size (); ++i)   {
int v = pre[i];
res = min (res, dp
[1][v]);
}

//debug
for (int i=1; i<=n; ++i)    {
printf ("%d ", dp[i][0][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=1; i<=n1; ++i)   G[i].clear ();
for (int i=1; i<=n1; ++i)   con[i][i] = true;
for (int i=1; 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=1; i<=n; ++i)    {
scanf ("%d", &a[i]);
}

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

return 0;
}