HDU 3472 HS BDC （混合图的欧拉路径判断）

HS BDC

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 748    Accepted Submission(s): 290

Problem Description

IELTS is around the corner! love8909 has registered for the exam, but he still hasn’t got prepared. Now he decides to take actions. But when he takes out the New Oriental IELTS Vocabulary, he finds there are so many words. But love8909 doesn’t get scared, because he has got a special skill. If he can make a list with some meaningful words, he will quickly remember these words and will not forget them. If the last letter of some word Wa is the same as the first letter of some word Wb, then you can connect these two words and make a list of two words. If you can connect a word to a list, you will make a longer list.

While love8909 is making the list, he finds that some words are still meaningful words if you reverse them. For example, if you reverse the word “pat”, you will get another meaningful word “tap”.

After scanning the vocabulary, love8909 has found there are N words, some of them are meaningful if reversed, while others are not. Now he wonders whether he can remember all these words using his special skill.

The N-word list must contain every word once and only once.

 

 

Input

An integer T (T <= 50) comes on the first line, indicating the number of test cases.

On the first line of each test cases is an integer N (N <= 1000), telling you that there are N words that love8909 wants to remember. Then comes N lines. Each of the following N lines has this format: word type. Word will be a string with only ‘a’~’z’, and type will be 0(not meaningful when reversed) or 1(meaningful when reversed). The length of each word is guaranteed to be less than 20.

 

 

Output

The format of the output is like “Case t: s”, t is the number of the test cases, starting from 1, and s is a string.
For each test case, if love8909 can remember all the words, s will be “Well done!”, otherwise it’s “Poor boy!”

 

 

Sample Input

3 6 aloha 0 arachnid 0 dog 0 gopher 0 tar 1 tiger 0 3 thee 1 earn 0 nothing 0 2 pat 1 acm 0

 

 

Sample Output

Case 1: Well done! Case 2: Well done! Case 3: Poor boy!

Hint

In the first case, the word “tar” is still meaningful when reversed, and love8909 can make a list as “aloha-arachnid-dog-gopher-rat-tiger”. In the second case, the word “thee” is still meaningful when reversed, and love8909 can make a list as “thee-earn-nothing”. In the third case, no lists can be created.

 

 

Author

allenlowesy

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（4）——Host by UESTC

 

 

 

 

 

几乎就是模板题了。

先要判断连通。

然后转化成网络流判断欧拉回路。

/* ***********************************************
Author        :kuangbin
Created Time  :2014-2-3 15:00:40
File Name     :E:\2014ACM\专题学习\图论\欧拉路\混合图\HDU3472.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 30;
//最大流部分
const int MAXM = 10000;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0)
{
    edge[tol].to = v;
    edge[tol].cap = w;
    edge[tol].next = head[u];
    edge[tol].flow = 0;
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = rw;
    edge[tol].next = head[v];
    edge[tol].flow = 0;
    head[v] = tol++;
}
int sap(int start,int end,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            for(int i = pre[u];i != -1;i = pre[edge[i^1].to])
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u]; i != -1;i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1;i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])
            {
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            u = v;
            continue;
        }
        
        int Min = N;
        for(int i = head[u];  i != -1;i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]] --;
        if(!gap[dep[u]])return ans;
        dep[u] = Min+1;
        gap[dep[u]]++;
        if(u != start) u = edge[pre[u]^1].to;
    }
    return ans;
}

int in[30],out[30];
int F[30];
int find(int x)
{
    if(F[x] == -1)return x;
    else return F[x] = find(F[x]);
}
void bing(int u,int v)
{
    int t1 = find(u), t2 = find(v);
    if(t1 != t2)F[t1] = t2;
}
char str[100];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T,n;
    scanf("%d",&T);
    int iCase = 0;
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        memset(F,-1,sizeof(F));
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        init();
        int k;
        int s = -1;
        while(n--)
        {
            scanf("%s%d",str,&k);
            int len = strlen(str);
            int u = str[0] - 'a';
            int v = str[len-1] - 'a';
            out[u]++;
            in[v]++;
            s = u;
            if(k == 1)
                addedge(u,v,1);
            bing(u,v);
        }
        bool flag = true;
        int cnt = 0;
        int s1 = -1, s2 = -1;
        for(int i = 0;i < 26;i++)
            if(in[i] || out[i])
            {
                if(find(i) != find(s))
                {
                    flag = false;
                    break;
                }
                if((in[i] + out[i])&1)
                {
                    cnt++;
                    if(s1 == -1)s1 = i;
                    else s2 = i;
                }
            }
        if(cnt != 0 && cnt != 2)flag = false;
        if(!flag)
        {
            printf("Case %d: Poor boy!\n",iCase);
            continue;
        }
        if(cnt == 2)
        {
            out[s1]++;
            in[s2]++;
            addedge(s1,s2,1);
        }
        for(int i = 0;i < 26;i++)
        {
            if(out[i] - in[i] > 0)
                addedge(26,i,(out[i] - in[i])/2);
            else if(in[i] - out[i] > 0)
                addedge(i,27,(in[i] - out[i])/2);
        }
        sap(26,27,28);
        for(int i = head[26];i != -1;i = edge[i].next)
            if(edge[i].cap > 0 && edge[i].cap > edge[i].flow)
            {
                flag = false;
                break;
            }
        if(flag)printf("Case %d: Well done!\n",iCase);
        else printf("Case %d: Poor boy!\n",iCase);
    }
    return 0;
}