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2016 ACM/ICPC Asia Regional Qingdao Online HDU5883
链接:http://acm.hdu.edu.cn/showproblem.php?pid=5883
解法:先判断是不是欧拉路,然后枚举
#pragma comment(linker, "/STACK:102400000,102400000")#include <math.h>#include <time.h>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <set>#include <map>#include <string>#include <stack>#include <queue>#include <vector>#include <bitset>#include <iostream>#include <algorithm>#define pb push_back#define fi first#define se second#define icc(x) (1<<(x))#define lcc(x) (1ll<<(x))#define lowbit(x) (x&-x)#define debug(x) cout<<#x<<"="<<x<<endl#define rep(i,s,t) for(int i=s;i<t;++i)#define per(i,s,t) for(int i=t-1;i>=s;--i)#define mset(g, x) memset(g, x, sizeof(g))using namespace std;typedef long long ll;typedef unsigned long long ull;typedef unsigned int ui;typedef double db;typedef pair<int,int> pii;typedef pair<ll,ll> pll;typedef vector<int> veci;const int mod=(int)1e9+7,inf=0x3fffffff,rx[]={-1,0,1,0},ry[]={0,1,0,-1};const ll INF=1ll<<60;const db pi=acos(-1),eps=1e-8;template<class T> void rd(T &res){ res = 0; int ch,sign=0; while( (ch=getchar())!=‘-‘ && !(ch>=‘0‘&&ch<=‘9‘)); if(ch == ‘-‘) sign = 1; else res = ch-‘0‘; while((ch=getchar())>=‘0‘&&ch<=‘9‘) res = (res<<3)+(res<<1)+ch-‘0‘; res = sign?-res:res;}template<class T>void rec_pt(T x){ if(!x)return; rec_pt(x/10); putchar(x%10^48);}template<class T>void pt(T x){ if(x<0) putchar(‘-‘),x=-x; if(!x)putchar(‘0‘); else rec_pt(x);}template<class T>inline void ptn(T x){ pt(x),putchar(‘\n‘); }template<class T>inline void Max(T &a,T b){ if(b>a)a=b; }template<class T>inline void Min(T &a,T b){ if(b<a)a=b; }template<class T>inline T mgcd(T b,T d){ return b?mgcd(d%b,b):d; }//gcd模板,传入的参数必须是用一类型//-------------------------------主代码--------------------------------------//int g[100100];int d[100100];int markpath[100100];int mark[100100];struct node{ int to,next;}edge[1000100];int cnt,pre[100100];void add_edge(int u,int v){ edge[cnt].to = v; edge[cnt].next = pre[u]; pre[u] = cnt++;}void dfs(int s){ mark[s] = 1; for(int p=pre[s];p!=-1;p=edge[p].next){ int v = edge[p].to; if(mark[v] == 1) continue; dfs(v); }}int main(){ int T; rd(T); while (T--) { cnt = 0; mset(d, 0); mset(markpath, 0); mset(pre, -1); mset(mark, 0); int n,m; rd(n),rd(m); for(int i=1;i<=n;i++) { rd(g[i]); } rep(i, 0, m){ int x,y; rd(x),rd(y); markpath[x] = 1; markpath[y] = 1; add_edge(x,y); add_edge(y,x); d[x]++; d[y]++; } int cnt = 0; for(int i=1;i<=n;i++){ if(d[i]%2 != 0) cnt++; } if(cnt!=0 && cnt!=2){ printf("Impossible\n"); continue; } int flag = 0; for(int i=1;i<=n;i++){ if(markpath[i]==1 && mark[i] ==0){ flag ++; dfs(i); } } if(flag > 1) { printf("Impossible\n"); continue; } int ans = 0; if(cnt == 0){ int sum = 0; for(int i=1;i<=n;i++){ if( (d[i]/2)%2!=0 ) { sum ^= g[i]; } } for(int i=1;i<=n;i++){ if(markpath[i]==1) ans = max(ans,sum^g[i]); } }else{ int sum = 0; for(int i=1;i<=n;i++){ if(d[i]%2!=0){ d[i]++; } if( (d[i]/2)%2!=0 ) { sum ^= g[i]; } } ans = sum; } ptn(ans); } return 0;}
2016 ACM/ICPC Asia Regional Qingdao Online HDU5883
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