POJ1330Nearest Common Ancestors最近公共祖先LCA问题

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用的离线算法Tarjan

该算法的详细解释请戳

http://www.cnblogs.com/Findxiaoxun/p/3428516.html

做这个题的时候，直接把1470的代码copy过来，改了改输入输出。这个的难度比那个低。

#include<cstdio>#include<algorithm>#include<cstring>#include<vector>using namespace std;const int MAXN=10005;int father[MAXN],ancestor[MAXN];bool visit[MAXN];int ans[MAXN];vector<int> map[MAXN];//save the treeint n,t,root,sx,sy;bool indegree[MAXN];//the indegree to find the rootint getfather(int v){//path compression    if(father[v]==v)return v;    return father[v]=getfather(father[v]);}void aunion(int u,int v){    int fv=getfather(v),fu=getfather(u);    father[fv]=fu;}bool isit(int a,int b){    if(a==sx&&b==sy)return true;    return false;}void LCA(int id){    int len=map[id].size();    int son;    for(int i=0;i<len;i++){        son=map[id][i];        LCA(son);        aunion(id,son);    }    visit[id]=1;    if(visit[sy]&&id==sx){        printf("%d\n",father[getfather(sy)]);        return;    }else{//attention:this way        if(visit[sx]&&id==sy){            printf("%d\n",father[getfather(sx)]);            return;        }    }}void init(){    int x,y,z;    scanf("%d",&n);    //initialize all the vars    for(int i=0;i<=n;i++){        map[i].clear();    }    memset(visit,0,sizeof(visit));    memset(ans,0,sizeof(ans));    memset(indegree,0,sizeof(indegree));    for(int i=0;i<=n;i++)father[i]=i;    for(int i=0;i<n-1;i++){        scanf("%d%d",&x,&y);        indegree[y]=1;        map[x].push_back(y);    }    scanf("%d%d",&sx,&sy);    for(int i=1;i<=n;i++)if(!indegree[i])root=i;//find the root;warning:the 0}int main(){    int  t;    scanf("%d",&t);    while(t--){        init();        LCA(root);    }    return 0;}

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