标签：hs bdc   hdu 3472   混合图的欧拉回路   

HS BDC

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

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.
 

题意：给出n个单词，问这些单词能否首尾相连组成一个整体，其中部分单词可以翻转。

思路：将每个单词的首尾字母看成结点，每个单词看成一条边，可以看出这是一张混合图，要求判断是否有欧拉路。使用判断欧来回路的方法，如果存在欧拉回路那么就一定存在欧拉路了；否则如果有且仅有两个点的（出度-入度）是奇数，那么给这两个点加边，判断是否存在欧拉回路。注意这一题要判断图的连通性。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 30
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
const int MAXM = 30000;
typedef long long ll;
using namespace std;

struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];

int st,ed;
int d[MAXN];
int tol,m,s;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
int in[MAXN],out[MAXN];

void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
}

//加边，单向图三个参数，双向图四个参数
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++;
}

//输入参数：起点，终点，点的总数
//点的编号没有影响，只要输入点的总数
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 father[MAXN];

int find_father(int x)
{
    if (x!=father[x])
        father[x]=find_father(father[x]);
    return father[x];
}

void Union(int a,int b)
{
    int fa=find_father(a);
    int fb=find_father(b);
    if (fa!=fb)
        father[fa]=fb;
}

int main()
{
//    freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
    int i,j,t,n,d,cas=1,x;
    char str[30];
    scanf("%d",&t);
    while (t--)
    {
        init();
        for (i=0;i<30;i++)
            father[i]=i;
        scanf("%d",&n);
        for (i=0;i<n;i++)
        {
            scanf("%s %d",str,&d);
            int len=strlen(str);
            out[str[0]-'a']++;
            in[str[len-1]-'a']++;
            x=str[0]-'a';
            if (d==1)
                addedge(str[0]-'a',str[len-1]-'a',1);
            Union(str[0]-'a',str[len-1]-'a');
        }
        int cnt=0;
        int w[30];
        bool flag=true;
        for (i=0;i<26;i++)
        {
            if (in[i]||out[i])
            {
                if (find_father(i)!=find_father(x))
                {
                    flag=false;
                    break;
                }
                if ((out[i]-in[i])%2)
                    w[cnt++]=i;
            }
        }
        if (cnt!=0&&cnt!=2||!flag)
        {
            printf("Case %d: Poor boy!\n",cas++);
            continue;
        }
        if (cnt==2)
        {
            in[w[0]]++;
            out[w[1]]++;
            addedge(w[0],w[1],1);
        }
        sap(26,27,28);
        for (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",cas++);
        else printf("Case %d: Poor boy!\n",cas++);
    }
    return 0;
}

HS BDC (hdu 3472 混合图的欧拉回路)

标签：hs bdc   hdu 3472   混合图的欧拉回路   

原文地址：http://blog.csdn.net/u014422052/article/details/45677353