hdu 5883 The Best Path 欧拉回路（路径）

The Best Path

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 924    Accepted Submission(s): 383

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

2 3 2 3 4 5 1 2 2 3 4 3 1 2 3 4 1 2 2 3 2 4

 

Sample Output

2 Impossible

 

题意：有一个图，每个点有一个权值，自己定一个起点，然后找一条路径，要求经过所有的边且每条边只能经过一次（点可以经过任意多次）。要求路径中经过所有点的异或值最大。

解法：每条边经过一次，先考虑是欧拉回路或欧拉路径，达成这个条件要求图一定是连通图，欧拉回路每个点的度数是偶数，欧拉路径要求有且只有2个点度数是奇数，其他点度数都为偶数。

针对欧拉路径的情况，因为起点终点已经固定是度数为奇数的点，画一下可以发现每个点经过的次数就是每个点度数除2向上取整（du[i]+1）/2。因为是异或，异或2次相当于没有经过，所以只要留下经过次数是奇数次的点，用一个ans记录下来就好了。

欧拉回路的话，随便画个环也能发现，除了起点，每个点经过的次数都是度数/2次（因为度数都是偶数，除2向上取整不改变，所以可以借用欧拉路径的ans），起点要多经过一次。然后枚举一下起点就可以了。

至于连通性，瞎dfs一下就好了。

CODE

#include 
using namespace std;
typedef long long LL;
const int N = 1e5+10;
struct node{
    int v,next;
}E[N*30];
int n,m,top;
int du[N];   ///每个点的度数
int num[N];  ///每个点的价值
int head[N]; ///邻接表头结点
bool vis[N]; ///dfs判断连通性用

void Init()
{
    top = 0;
    for(int i = 0;i <= n;i++){
        du[i] = num[i] = 0;
        head[i] = -1;
        vis[i] = false;
    }
}

void add(int u,int v)
{
    E[top].v = v;
    E[top].next = head[u];
    head[u] = top++;
}

void dfs(int u)
{
    vis[u] = true;
    for(int i = head[u];i != -1;i = E[i].next){
        int v = E[i].v;
        if(vis[v]) continue;
        dfs(v);
    }
}

int main(void)
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        Init();
        for(int i = 1;i <= n;i++)
            scanf("%d",&num[i]);
        for(int i = 1;i <= m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            du[u]++;
            du[v]++;
            add(u,v);
            add(v,u);
        }
        dfs(1);  ///判断连通性
        bool flag = false;
        for(int i = 1;i <= n;i++){
            if(!vis[i]) flag = true;
        }
        if(flag){
            printf("Impossible\n");
            continue;
        }
        int cnt = 0;
        for(int i = 1;i <= n;i++){
            if(du[i]%2 == 1) cnt++;
        }
        if(cnt > 2){  ///度数为奇数的点个数大于2个，不存在欧拉回路和欧拉路径
            printf("Impossible\n");
            continue;
        }
        int ans = 0;
        for(int i = 1;i <= n;i++){  ///欧拉路径的情况
            du[i] = (du[i]+1)/2;
            if(du[i]%2 == 1)
                ans ^= num[i];
        }
        if(cnt == 0){   ///欧拉回路的话枚举下起点取最大值
            int tmp = ans;
            for(int i = 1;i <= n;i++){
                int t = tmp^num[i];
                ans = max(ans,t);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}