[并查集+混合图欧拉回路+网络流]hdoj 3472：HS BDC

大致题意：

    给出你n个单词，如果一个单词的尾部和另一个单词的头部相同那么这两个单词就能连在一起(比如‘abc’和‘cde’就能够连成abc-cde)，如果这个单词后面的数字是1，则代表这个单词的逆序也是一个单词，可以翻转过来。求所有单词能不能连成一长串。

 

大致思路：
    转化为混合图的欧拉回路，把字母当作节点。注意要用并查集来检测图是否连通。

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int inf=1<<30;
const int nMax=2000;
const int mMax=100000;

class node{
    public:
    int c,u,v,next;
};node edge[mMax];
int ne, head[nMax];
int cur[nMax], ps[nMax], dep[nMax];

void addedge(int u, int v,int w){   ////dinic邻接表加边
 //   cout<<u<<" add "<<v<<endl;
    edge[ne].u = u;
    edge[ne].v = v;
    edge[ne].c = w;
    edge[ne].next = head[u];
    head[u] = ne ++;
    edge[ne].u = v;
    edge[ne].v = u;
    edge[ne].c = 0;
    edge[ne].next = head[v];
    head[v] = ne ++;
}

int dinic(int s, int t){                       //  dinic
    int tr, res = 0;
    int i, j, k, f, r, top;
    while(1){
        memset(dep, -1, sizeof(dep));
        for(f = dep[ps[0]=s] = 0, r = 1; f != r;)
            for(i = ps[f ++], j = head[i]; j; j = edge[j].next)
                if(edge[j].c && dep[k=edge[j].v] == -1){
                    dep[k] = dep[i] + 1;
                    ps[r ++] = k;
                    if(k == t){
                        f = r; break;
                    }
                }
        if(dep[t] == -1) break;
        memcpy(cur, head, sizeof(cur));
        i = s, top = 0;
        while(1){
            if(i == t){
                for(tr =inf, k = 0; k < top; k ++)
                    if(edge[ps[k]].c < tr)
                        tr = edge[ps[f=k]].c;
                for(k = 0; k < top; k ++){
                    edge[ps[k]].c -= tr;
                    edge[ps[k]^1].c += tr;
                }
                i = edge[ps[top=f]].u;
                res += tr;
            }
            for(j = cur[i]; cur[i]; j = cur[i] =edge[cur[i]].next){
                if(edge[j].c && dep[i]+1 == dep[edge[j].v]) break;
            }
            if(cur[i]){
                ps[top ++] = cur[i];
                i = edge[cur[i]].v;
            }
            else{
                if(top == 0) break;
                dep[i] = -1;
                i = edge[ps[-- top]].u;
            }
        }
    }
    return res;
}

char str[30];
int in[nMax];
int out[nMax];
bool vis[nMax];

int p[nMax],r[nMax];
int find(int a){
    if(p[a]!=a){
        p[a]=find(p[a]);
    }
    return p[a];
}
void unionset(int i,int j){
    i=find(i);
    j=find(j);
    if(i!=j){
        if(r[i]>r[j]){
            p[j]=i;
        }
        else{
            p[i]=j;
            if(r[i]==r[j]){
                r[i]++;
            }
        }
    }
}

int main(){
    int i,j,k,cas,a,b,n,f,len,flag,s,t,sum;
    scanf("%d",&cas);
    for(int ii=1;ii<=cas;ii++){
        s=27,t=28;
        ne=2;
        flag=1;
        memset(head,0,sizeof(head));
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(vis,0,sizeof(vis));
        for(i=0;i<100;i++)p[i]=i,r[i]=0;
        scanf("%d",&n);
        for(i=0;i<n;i++){
            scanf("%s%d",str,&f);
            len=strlen(str);
            in[str[len-1]-'a']++;
            out[str[0]-'a']++;
            vis[str[0]-'a']=vis[str[len-1]-'a']=1;
            unionset(str[len-1]-'a',str[0]-'a');
            if(f==1){
                addedge(str[0]-'a',str[len-1]-'a',1);
            }
        }
        printf("Case %d:",ii);
        for(i=0;i<26;i++){
            if(vis[i]){
                for(j=i+1;j<26;j++){
                    if(vis[j]){
                        if(find(i)!=find(j)){
                            flag=0;
                            break;
                        }
                    }
                }
                break;
            }
        }
        int v1=-1,v2=-1,tmp=0;
        for(i=0;i<26;i++){
            if(!vis[i])continue;
            if((in[i]-out[i])%2==1){
                v1=i;
                tmp++;
            }
            if((in[i]-out[i])%2==-1){
                v2=i;
                tmp++;
            }
        }
        if(tmp==0||(tmp==2&&v1!=-1&&v2!=-1)){
            if(tmp==2){
                addedge(v1,v2,1);
                in[v2]++;
                out[v1]++;
            }
        }
        else{
            flag=0;
        }
        if(!flag){
            printf(" Poor boy!\n");
            continue;
        }
        sum=0;
        for(i=0;i<26;i++){
            if(!vis[i])continue;
            tmp=(out[i]-in[i])/2;
            if(tmp<0){
                addedge(i,t,-tmp);
            }
            else if(tmp>0){
                sum+=tmp;
                addedge(s,i,tmp);
            }
        }
        if(dinic(s,t)!=sum){
            printf(" Poor boy!\n");
        }
        else{
            printf(" Well done!\n");
        }
    }
    return 0;
}