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UVA 116 Unidirectional TSP DP
DP,对于输出字典序最小方案直接反着递推就好了。
#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 = 20;const int maxm = 200;int num[maxn][maxm], n, m;int f[maxn][maxm], pre[maxn][maxm];int main() { while(scanf("%d%d",&n,&m) != EOF) { for(int i = 0;i < n;i++) { for(int j = 0;j < m;j++) { scanf("%d",&num[i][j]); } } memset(f,0x3f,sizeof(f)); for(int i = 0;i < n;i++) f[i][m - 1] = num[i][m - 1]; for(int j = m - 1;j > 0;j--) { for(int i = 0;i < n;i++) { int c[3] = {i,(i + 1) % n,(i - 1 + n) % n}; for(int k = 0;k < 3;k++) { if(f[c[k]][j - 1] > f[i][j] + num[c[k]][j - 1]) { f[c[k]][j - 1] = f[i][j] + num[c[k]][j - 1]; pre[c[k]][j - 1] = i; } } } } int ans = INF; for(int i = 0;i < n;i++) ans = min(ans,f[i][0]); for(int i = 0;i < n;i++) if(ans == f[i][0]) { int nowrow = i; printf("%d",nowrow + 1); for(int j = 1;j < m;j++) { nowrow = pre[nowrow][j - 1]; printf(" %d",nowrow + 1); } break; } printf("\n%d\n",ans); } return 0;}
UVA 116 Unidirectional TSP DP
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