[hdu5418 Victor and World]floyd + 状压DP 或 SPFA

题意：给n个点，m条边，每次只能沿边走，花费为边权值，求从1出发经过所有其它点≥1次最后回到1的最小花费。

思路：

1. 状压DP。先用Floyd得到任意两点间的最短距离，转移时沿两个点的最短路转移。此时的状态表示为dp[i][j]：“落脚点集合为i，最后停在j”的方案数；而不是“访问过的点的集合为i，最后停在j”的方案数。
2. SPFA。每个状态保存访问过的点的集合，以及最后停留的位置，然后SPFA即可。

floyd + 状压DP

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#pragma comment(linker, "/STACK:10240000") #include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define fillarray(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; #ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} #endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} const double PI = acos(-1.0); const int INF = 0x3f3f3f3f; const double EPS = 1e-12; /* -------------------------------------------------------------------------------- */ const int maxn = ; int e[maxn][maxn], dist[maxn][maxn], dp[ << maxn][maxn]; int n; void add(int u, int v, int w) { umin(e[u][v], w); } void work() { fillchar(dp, 0x3f); for (int i = ; i < n; i ++) { for (int j = ; j < n; j ++) { dist[i][j] = e[i][j]; } } for (int k = ; k < n; k ++) { for (int i = ; i < n; i ++) { for (int j = ; j < n; j ++) { if (k != i && k != j && i != j) { umin(dist[i][j], dist[i][k] + dist[k][j]); } } } } dp[][] = ; for (int i = ; i < ( << n); i ++) { for (int j = ; j < n; j ++) { if (i & ( << j)) { for (int k = ; k < n; k ++) { if (k != j) if (i & ( << k)) { umin(dp[i][j], dp[i ^ ( << j)][k] + dist[k][j]); } } } } } int ans = dp[( << n) - ][]; for (int i = ; i < n; i ++) { umin(ans, dp[( << n) - ][i] + dist[i][]); } cout << ans << endl; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T, u, v, w, m; cin >> T; while (T --) { cin >> n >> m; fillchar(e, 0x3f); for (int i = ; i < m; i ++) { scanf("%d%d%d", &u, &v, &w); u --; v --; add(u, v, w); add(v, u, w); } work(); } return ; }

SPFA:

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#pragma comment(linker, "/STACK:10240000") #include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define fillarray(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; #ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} #endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} const double PI = acos(-1.0); const int INF = 0x3f3f3f3f; const double EPS = 1e-12; /* -------------------------------------------------------------------------------- */ const int maxn = ; int n; int e[maxn][maxn], d[maxn][ << maxn], vis[maxn][ << maxn]; void add(int u, int v, int w) { u --; v --; umin(e[u][v], w); } bool relax(pii u, pii v, int w) { if (d[v.X][v.Y] > d[u.X][u.Y] + w) { d[v.X][v.Y] = d[u.X][u.Y] + w; return true; } return false; } void work() { fillchar(vis, ); fillchar(d, 0x3f); queue<pii> Q; Q.push(mp(, )); d[][] = ; vis[][] = true; while (!Q.empty()) { pii H = Q.front(); Q.pop(); vis[H.X][H.Y] = false; for (int i = ; i < n; i ++) { if (e[H.X][i] < INF) { pii P = mp(i, H.Y | ( << i)); if (relax(H, P, e[H.X][i]) && !vis[P.X][P.Y]) { vis[P.X][P.Y] = true; Q.push(P); } } } } cout << d[][( << n) - ] << endl; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T, u, v, w, m; cin >> T; while (T --) { cin >> n >> m; fillchar(e, 0x3f); for (int i = ; i < m; i ++) { scanf("%d%d%d", &u, &v, &w); add(u, v, w); add(v, u, w); } work(); } return ; }