• 【UVa】Salesmen(dp)


    http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4170

    d[i,j]前i个字符j结尾

    d[i,j]=min{d[i-1,k], j和k有边}+(a[i]!=j)

    初始化d[1,j]=1, d[1,a[1]]=0

    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <string>
    #include <iostream>
    #include <algorithm>
    #include <queue>
    #include <set>
    #include <map>
    using namespace std;
    typedef long long ll;
    #define pii pair<int, int>
    #define mkpii make_pair<int, int>
    #define pdi pair<double, int>
    #define mkpdi make_pair<double, int>
    #define pli pair<ll, int>
    #define mkpli make_pair<ll, int>
    #define rep(i, n) for(int i=0; i<(n); ++i)
    #define for1(i,a,n) for(int i=(a);i<=(n);++i)
    #define for2(i,a,n) for(int i=(a);i<(n);++i)
    #define for3(i,a,n) for(int i=(a);i>=(n);--i)
    #define for4(i,a,n) for(int i=(a);i>(n);--i)
    #define CC(i,a) memset(i,a,sizeof(i))
    #define read(a) a=getint()
    #define print(a) printf("%d", a)
    #define dbg(x) cout << (#x) << " = " << (x) << endl
    #define error(x) (!(x)?puts("error"):0)
    #define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
    #define printarr1(a, b) for1(_, 1, b) cout << a[_] << '	'; cout << endl
    inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
    inline const int max(const int &a, const int &b) { return a>b?a:b; }
    inline const int min(const int &a, const int &b) { return a<b?a:b; }
    
    const int N=505;
    int mp[N][N], a[N], len, d[N][N], n, m;
    void add(int u, int v) { mp[u][v]=mp[v][u]=1; }
    int main() {
    	int cs=getint();
    	while(cs--) {
    		read(n); read(m);
    		CC(mp, 0); CC(d, 0x3f);
    		for1(i, 1, n) add(i, i);
    		for1(i, 1, m) add(getint(), getint());
    		read(len);
    		for1(i, 1, len) read(a[i]);
    		for1(i, 1, n) d[1][i]=1;
    		d[1][a[1]]=0;
    		for1(i, 2, len) for1(j, 1, n) for1(k, 1, n) if(mp[j][k]) d[i][j]=min(d[i][j], d[i-1][k]+(a[i]!=j));
    		int ans=~0u>>1;
    		for1(i, 1, n) ans=min(ans, d[len][i]);
    		printf("%d
    ", ans);
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/iwtwiioi/p/4069288.html
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