hdu 3472 HS BDC(混合路的欧拉路径)

这题是混合路的欧拉路径问题。

1.判断图的连通性，若不连通，无解。

2.给无向边任意定向，计算每个结点入度和出度之差deg[i]。deg[i]为奇数的结点个数只能是0个或2个，否则肯定无解。

3.(若存在2个deg[i]为奇数的结点，则在两点连一条流量为1的边，方向任意)设立源点s和汇点t（自己另外定两个点），若某点i入度<出度，连边(s,i,-deg[i]/2)，若入度>出度，连边(i,t,deg[i]/2)；对于任意定向的无向边(i,j,1)。

5.若从S发出的边全部满流，证明存在欧拉回路(路径)，否则不存在。

view code#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 1<<30;
const int N = 2010;
int _,cas=1, n, fa[30], pre[30], deg[30], s, t, d[30];
int cur[30];
bool vis[30];

struct edge
{
    int u, v, cap, next;
    edge() {}
    edge(int u, int v, int w, int next):u(u),v(v),cap(w),next(next) {}
}e[N<<1];
int ecnt;

void addedge(int u, int v, int w)
{
    e[ecnt] = edge(u, v, w, pre[u]);
    pre[u] = ecnt++;
    e[ecnt] = edge(v, u, 0, pre[v]);
    pre[v] = ecnt++;
}

int find(int x)
{
    return x==fa[x]?x:(fa[x]=find(fa[x]));
}

void Union(int u, int v)
{
    u = find(u), v= find(v);
    if(u==v) return ;
    fa[u] = v;
}

bool is_ok()
{
    for(int i=0; i<26; i++) if(vis[i] && find(i)!=find(s)) return false;
    int res = 0;
    for(int i=0; i<26; i++) if(deg[i]%2){
        res++;
        if(res==1) s = i;
        else t = i;
    }
    return res==0 || res==2;
}

bool BFS()
{
    memset(vis, 0, sizeof(vis));
    queue<int >q;
    q.push(s);
    d[s] = 0, vis[s]=1;
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        for(int i=pre[u]; ~i; i=e[i].next)
        {
            int v = e[i].v;
            if(!vis[v] && e[i].cap>0)
            {
                vis[v] = 1;
                d[v] = d[u] +1;
                q.push(v);
            }
        }
    }
    return vis[t];
}

int DFS(int u, int c)
{
    if(u==t || c==0) return c;
    int flow=0, f;
    for(int &i=pre[u]; ~i; i=e[i].next)
    {
        int v = e[i].v;
        if(d[v]==d[u]+1 && (f=DFS(v, min(c, e[i].cap)))>0)
        {
            e[i].cap -= f;
            e[i^1].cap += f;
            flow += f;
            c -= f;
            if(c==0) break;
        }
    }
    return flow;
}

int Maxflow()
{
    int flow = 0;
    while(BFS())
    {
        memcpy(cur, pre, sizeof(pre));
        flow += DFS(s, INF);
    }
    return flow;
}

void solve()
{
    scanf("%d", &n);
    memset(pre, -1, sizeof(pre));
    memset(deg, 0, sizeof(deg));
    memset(vis, 0, sizeof(vis));
    for(int i=0; i<27; i++) fa[i] = i;
    ecnt = 0;
    int rev; char str[27];
    for(int i=0; i<n; i++)
    {
        scanf("%s%d", str, &rev);
        int u = str[0]-'a',  v = str[strlen(str)-1]-'a';
        deg[u]--; deg[v]++;
        vis[u] = vis[v] = true;
        s = u;
        if(rev) addedge(u, v, 1);
        Union(u,v);
    }
    t = -1;
    printf("Case %d: ", cas++);
    if(!is_ok()) {
        puts("Poor boy!");
        return ;
    }
    if(t!=-1) addedge(t, s, 1);
    s = 26, t = 27;
    int fulflow = 0, flow;
    for(int i=0; i<26; i++)
    {
        if(deg[i]==0) continue;
        if(deg[i]<0) addedge(s, i, -deg[i]/2);
        else addedge(i, t, deg[i]/2), fulflow+=deg[i]/2;
    }
    flow = Maxflow();
    puts(flow==fulflow?"Well done!":"Poor boy!");
}

int main()
{
//    freopen("in.txt", "r", stdin);
    cin>>_;
    while(_--) solve();
    return 0;
}