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_back
template<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_back
template<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;
}