首页 > 代码库 > 2016青岛网络赛 The Best Path

2016青岛网络赛 The Best Path

2024-08-13 16:42:55   226人阅读

The Best Path

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Problem Description
Alice is planning her travel route in a beautiful valley. In this valley, there are N lakes, and M rivers linking these lakes. Alice wants to start her trip from one lake, and enjoys the landscape by boat. That means she need to set up a path which go through every river exactly once. In addition, Alice has a specific number (a1,a2,...,an) for each lake. If the path she finds is P0→P1→...→Pt, the lucky number of this trip would be aP0XORaP1XOR...XORaPt. She want to make this number as large as possible. Can you help her?
 

 

Input
The first line of input contains an integer t, the number of test cases. t test cases follow.

For each test case, in the first line there are two positive integers N (N≤100000) and M (M≤500000), as described above. The i-th line of the next N lines contains an integer ai(∀i,0≤ai≤10000) representing the number of the i-th lake.

The i-th line of the next M lines contains two integers ui and vi representing the i-th river between the ui-th lake and vi-th lake. It is possible that ui=vi.
 

 

Output
For each test cases, output the largest lucky number. If it dose not have any path, output "Impossible".
 

 

Sample Input
23 23451 22 34 312341 22 32 4
 

 

Sample Output
2 Impossible
分析：先判断有几个顶点数>1的联通块，如果没有，取最大值即可，如果>1个，不可能；
　　　否则判断是欧拉路还是欧拉回路，欧拉路每个点直接求贡献即可，回路则要取一个使得答案最大的起点，因为起点多算1次；
代码：
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 1000000007#define inf 0x3f3f3f3f#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=1e5+10;using namespace std;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int n,m,k,t,p[maxn],a[maxn],d[maxn],q[maxn],ans;int find(int x){    return p[x]==x?x:p[x]=find(p[x]);}int main(){    int i,j;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        ans=0;        memset(d,0,sizeof d);        memset(q,0,sizeof q);        rep(i,1,n)scanf("%d",&a[i]),p[i]=i;        while(m--)        {            int b,c;            scanf("%d%d",&b,&c);            d[b]++,d[c]++;            int fa=find(b),fb=find(c);            if(fa!=fb)p[fa]=fb;        }        int cnt=0;        rep(i,1,n)        {            int fa=find(i);            if(p[fa]!=i&&!q[p[fa]])cnt++,q[p[fa]]=1;        }        if(cnt>1){puts("Impossible");continue;}        else if(cnt==0)        {            rep(i,1,n)ans=max(ans,a[i]);            printf("%d\n",ans);            continue;        }        cnt=0;        rep(i,1,n)        {            if(d[i]&1)cnt++;            int now=(d[i]+1)/2;            if(now&1)ans^=a[i];        }        if(cnt==2)        {            printf("%d\n",ans);        }        else if(cnt==0)        {            ans=0;            int ma=0;            rep(i,1,n)            {                int now=(d[i]+1)/2;                if(now&1)ans^=a[i];            }            rep(i,1,n)if(d[i])ma=max(ma,ans^a[i]);            printf("%d\n",ma);        }        else puts("Impossible");    }    //system("Pause");    return 0;}

2016青岛网络赛 The Best Path


联系
我们