Hdu 5348 MZL's endless loop (构造)

来源：互联网发布：投资域名编辑：程序博客网时间：2024/06/16 01:29

解析：

每找到一个单个的环，将其指定顺序后从图中出，最后会得到一个森林，对于每一棵树，从根节点开始根据出入大小指定方向即可。注意实现细节！

[code]:

#include<cstdio>#include<cstring>#include<algorithm>#include<stack>#include<queue>using namespace std;typedef pair<int,int> P;const int maxn = 1e5+5;const int maxm = 3e5+5;struct Nod{    int b,next,id;    void init(int b,int next,int id){        this->b=b;this->next=next;this->id=id;    }}buf[2*maxn];int n,m,len,E[maxn],ans[maxm],del[maxm],instack[maxn],used[maxn];int out[maxn],in[maxn];queue<P> G[maxn];void init(){    len = 0;    memset(E,-1,(n+1)*sizeof(int));    memset(del,0,m*sizeof(int));    memset(used,0,(n+1)*sizeof(int));    memset(out,0,(n+1)*sizeof(int));    memset(in,0,(n+1)*sizeof(int));    for(int i = 1;i <= n;i++) while(!G[i].empty()) G[i].pop();}void add_edge(int a,int b,int c){    buf[len].init(b,E[a],c);E[a]=len++;    buf[len].init(a,E[b],c^1);E[b]=len++;}int findCircle(int u){    int i,v,c;    if(instack[u]) return u;    instack[u] = 1;    while(!G[u].empty()){        P p = G[u].front();G[u].pop();        v = p.first;i = p.second;        if(del[i/2]) continue;        del[i/2] = 1;        c = findCircle(v);        if(c != -1&&c != u){            ans[i/2] = i&1;            instack[u] = 0;            return c;        }else if(c == u) ans[i/2] = i&1;        else add_edge(u,v,i);    }    instack[u] = 0;    return -1;}void dfs(int u,int pre){    int i,v,id;    used[u] = 1;    for(i = E[u];i != -1;i = buf[i].next){        v = buf[i].b;id = buf[i].id;        if(v == pre) continue;        if(out[u]>in[u]) ans[id/2]=1^(id&1),in[u]++,out[v]++;        else ans[id/2] = 0^(id&1),out[u]++,in[v]++;        dfs(v,u);    }}int main(){    int i,j,cas,u,v;    scanf("%d",&cas);    while(cas--){        scanf("%d%d",&n,&m);        init();        for(i = 0;i < m;i++){            scanf("%d%d",&u,&v);            G[u].push(P(v,2*i));            G[v].push(P(u,2*i+1));        }        for(i = 1;i <= n;i++) findCircle(i);        for(i = 1;i <= n;i++){            if(used[i]) continue;            dfs(i,-1);        }        for(i = 0;i < m;i++) printf("%d\n",ans[i]);    }    return 0;}

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