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\n", ans);

    }

    return 0;

}