题意: n
个点 m
条无向边的图,找一个欧拉通路/回路使得这个路径所有结点的异或值最大。
析:由欧拉路性质,奇度点数量为0或2。一个节点被进一次出一次,度减2,产生一次贡献,因此节点 i 的贡献为 i 点的度数除以2然后再模22??degree?u????⌋ mod 2)∗a?u??
。欧拉回路的起点贡献多一次,
欧拉通路的起点和终点贡献也多一次。因此如果是欧拉回路的话枚举一下起点就好了。
但是这个题有坑,就是有孤立点,这些点可以不连通,。。。。被坑死了,就是这一点,到最后也没过。。。伤心
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 5; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int p[maxn], in[maxn]; int Find(int x) { return x == p[x] ? x : p[x] = Find(p[x]); } int a[maxn]; int cnt; vector<int> vv; bool judge(){ int x = Find(1); cnt = 0;vv.clear(); for(int i = 1; i <= n; ++i){ if(x != Find(i) && i != Find(i)) return false; if(in[i] & 1) ++cnt, vv.push_back(i); if(cnt > 2) return false; } if(cnt && cnt != 2) return false; return true; } int main(){ int T; cin >> T; while(T--){ scanf("%d %d", &n, &m); for(int i = 1; i <= n; ++i){ p[i] = i; scanf("%d", &a[i]); } memset(in, 0, sizeof in); int u, v; for(int i = 0; i < m; ++i){ scanf("%d %d", &u, &v); int x = Find(u); int y = Find(v); if(x != y) p[y] = x; ++in[u]; ++in[v]; } if(!m){ printf("0\n"); continue; } if(!judge()){ printf("Impossible\n"); continue; } int ans = 0; for(int i = 1; i <= n; ++i){ int t = in[i]/2; if(t & 1) ans ^= a[i]; } if(cnt){ ans ^= a[vv[0]]; ans ^= a[vv[1]]; } else{ int x = ans; for(int i = 1; i <= n; ++i){ if(ans < (x ^ a[i])){ ans = x ^ a[i]; } } } printf("%d\n", ans); } return 0; }
时间: 2024-10-29 10:46:57