UVaLive 4256 Salesmen (简单DP)

题意：给一个无向连通图，和一个序列，修改尽量少的数，使得相邻两个数要么相等，要么相邻。

析：dp[i][j] 表示第 i 个数改成 j 时满足条件。然后就很容易了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn<<1][maxn];
bool G[maxn][maxn];
int a[maxn<<1];

int main(){
    int T;  cin >> T;
    while(T--){
        int t;
        scanf("%d %d", &n, &m);
        memset(G, false, sizeof G);
        for(int i = 0; i < m; ++i){
            int u, v;
            scanf("%d %d", &u, &v);
            G[u][v] = G[v][u] = true;
        }
        scanf("%d", &t);
        for(int i = 0; i < t; ++i)  scanf("%d", a+i);
        memset(dp, INF, sizeof dp);
        for(int i = 1; i <= n; ++i)  dp[0][i] = a[0] == i ? 0 : 1;
        for(int i = 1; i < t; ++i)
            for(int j = 1; j <= n; ++j){
                dp[i][j] = a[i] == j ? dp[i-1][j] : dp[i-1][j] + 1;
                for(int k = 1; k <= n; ++k)
                    if(G[j][k])   dp[i][j] = min(dp[i][j], a[i] == j ? dp[i-1][k] : dp[i-1][k] + 1);
            }

        int ans = INF;
        for(int i = 1; i <= n; ++i)  ans = min(ans, dp[t-1][i]);
        printf("%d
", ans);
    }
    return 0;
}