lightoj1086 Jogging Trails

有个无向网络，现在Robin想从某点出发，经历每条边至少一次，最后回到原点，求最少的权和。

这个有点像是欧拉回路，其实就是的，只是呢，，，有的边会许会重复走。在欧拉回路中，点的度数必然是偶数，这题中的度数为奇数的点的偶数也必然是偶数个，因为这个是无向图。那么，图是连通的，所以最后我们需要将度数为奇数的点进行建边，其实就是这两点之间的最短路上的边再走一次。点只有15个，，，所以状压。dp[sta]表示偶数点的状态为sta的时候的最短路（回路）。从最开始的状态枚举到终态，也就是dp[(1<<n) - 1]。

// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;// #define DEBUG#ifdef DEBUG#define debug(...) printf( __VA_ARGS__ )#else#define debug(...)#endif#define CLR(x) memset(x, 0,sizeof x)#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtemplate<class T> inline T Get_Max(const T&a,const T&b){return a < b?b:a;}template<class T> inline T Get_Min(const T&a,const T&b){return a < b?a:b;}typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;const double eps = 1e-10;const double PI = acos(-1.0);const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;int dist[20][20];int deg[20];int dp[(1<<15)+10];//dp[i]表示偶数定点状态为i时的最短路；int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);int t, n, m;int icase = 0;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);memset(dp, INF,sizeof dp);memset(dist, INF,sizeof dist);memset(deg, 0,sizeof deg);int u, v, c;int sum = 0, sta = 0;for (int i = 1;i <= m;++i){scanf("%d%d%d",&u,&v,&c);sum += c;deg[u]++,deg[v]++;dist[u][v] = dist[v][u] = min(dist[u][v], c);}for (int i = 1;i <= n;++i){dist[i][i] = 0;if (deg[i]%2==0){sta |= (1<<(i-1));}}for (int k = 1;k <= n;++k){for (int i = 1;i <= n;++i){for (int j = 1;j <= n;++j){dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);}}}dp[sta] = sum;//初始状态for (int i = sta;i < (1<<n)-1;++i){//枚举状态for (int j = 0;j < n;++j){if (i & (1<<j)) continue;//(j+1)节点此时的度数为奇；for (int k = 0;k < n;++k){if (j==k) continue;if (i & (1<<k)) continue;int now = (i | (1<<j));now |= (1<<k);//(k,j)建边dp[now] = min(dp[now],dp[i] + dist[j+1][k+1]);}}}printf("Case %d: %d\n", ++icase,dp[(1<<n)-1]);}return 0;}

这个跑得更快

// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;// #define DEBUG#ifdef DEBUG#define debug(...) printf( __VA_ARGS__ )#else#define debug(...)#endif#define CLR(x) memset(x, 0,sizeof x)#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtemplate<class T> inline T Get_Max(const T&a,const T&b){return a < b?b:a;}template<class T> inline T Get_Min(const T&a,const T&b){return a < b?a:b;}typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;const double eps = 1e-10;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;int dist[20][20];int deg[20];int dp[(1<<15)+10];int n,m;int dfs(int sta){if (dp[sta] != INF) return dp[sta];if (sta == (1<<n)-1) return 0;for (int i = 0;i < n;++i){if (!(sta & (1<<i))){for (int j = i + 1;j < n;++j){if (!(sta & (1<<j))){int nxt = sta|(1<<i)|(1<<j);dp[sta] = min(dp[sta], dfs(nxt) + dist[i+1][j+1]);}}}}return dp[sta];}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);int t, icase = 0;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);memset(dp, INF,sizeof dp);memset(dist, INF,sizeof dist);memset(deg, 0,sizeof deg);int sum = 0, sta = 0;int u, v, c;for (int i = 1;i <= m;++i){scanf("%d%d%d",&u,&v,&c);sum += c;deg[u]++;deg[v]++;dist[u][v] = dist[v][u] = min(dist[u][v], c);}for (int i = 1;i <= n;++i){dist[i][i] = 0;if (deg[i]%2==0){sta |= (1<<(i-1));}}for (int k = 1;k <= n;++k){for (int i = 1;i <= n;++i){for (int j = 1;j <= n;++j)dist[i][j] = min(dist[i][j],dist[i][k] + dist[k][j]);}}// dp[sta] = sum;sum += dfs(sta);printf("Case %d: %d\n", ++icase, sum);}return 0;}