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POJ 3311 Hie with the Pie TSP+Floyd
保证每个点访问过一次就行,然后会到原点。 这种情况可以先做一边floyd,然后跑tsp就好。
#include <cstdio>#include <cstring>#include <iostream>#include <map>#include <set>#include <vector>#include <string>#include <queue>#include <deque>#include <bitset>#include <list>#include <cstdlib>#include <climits>#include <cmath>#include <ctime>#include <algorithm>#include <stack>#include <sstream>#include <numeric>#include <fstream>#include <functional>using namespace std;#define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int,int> pii;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 15;int n,dis[maxn][maxn];int f[maxn][1 << maxn];int dfs(int now,int state) { if(state == (1 << (n + 1)) - 1) return dis[now][0]; if(f[now][state] != -1) return f[now][state]; int ret = INF; for(int i = 0;i <= n;i++) if(!(state & (1 << i))) { ret = min(ret,dfs(i,state | (1 << i)) + dis[now][i]); } return f[now][state] = ret;}int main() { while(scanf("%d",&n),n) { memset(f,-1,sizeof(f)); for(int i = 0;i <= n;i++) { for(int j = 0;j <= n;j++) { scanf("%d",&dis[i][j]); } } for(int k = 0;k <= n;k++) { for(int i = 0;i <= n;i++) { for(int j = 0;j <= n;j++) { dis[i][j] = min(dis[i][j],dis[i][k] + dis[k][j]); } } } int ret = dfs(0,1); printf("%d\n",ret); } return 0;}
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