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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