示例#1
0
static void Union(int x,int y)
{
	int fx = get_father(x);
	int fy = get_father(y);
	int ndisx = ndis[x];
	int ndisy = ndis[y];
#ifdef DBGMSG
	printf("union x : %d y : %d fx : %d fy : %d\n",x+1,y+1,fx+1,fy+1);
#endif
	if(fx == fy){
#ifdef DBGMSG
		printf("fx == fy!return\n");
#endif
		return;
	}
	if(rank[fx] > rank[fy]){
		father[fy] = fx;
		ndis[fy] = (ndisx + ndisy + 1)%2;
#ifdef DBGMSG
		printf("father[%d] == %d\n",fy+1,fx+1);
		printf("ndis[%d] = (ndis[%d] + ndis[%d] + 1)%2 = (%d + %d +1)%%2 = %d\n",fy+1,x+1,y+1,ndisx,ndisy,ndis[fy]);
#endif
	}else{
		father[fx] = fy;
		if(rank[fx] == rank[fy]){
			rank[fy]++;
		}
		ndis[fx] = (ndisx + ndisy + 1)%2;
#ifdef DBGMSG
		printf("father[%d] == %d\n",fx+1,fy+1);
		printf("ndis[%d] = (ndis[%d] + ndis[%d] + 1)%2 = (%d + %d +1)%%2 = %d\n",fx+1,x+1,y+1,ndisx,ndisy,ndis[fx]);
#endif
	}
	return;
}
示例#2
0
int main(){
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++){
		scanf("%d%d",&co[i][0],&co[i][1]);
		father[i]=i;
		flag[i]=false;
		for(j=1;j<i;j++){
			edge[i][j]=false;
			edge[j][i]=false;
			if (judge_distance(m,i,j)){
				edge[i][j]=true;
				edge[j][i]=true;
			}
		}
	}
	while(scanf("%c",&c)!=EOF){
		if (c=='O'){
			scanf("%d",&l);
			flag[l]=true;
			for(i=1;i<=n;i++){
				if ((flag[i])&&(edge[i][l])){
					father[get_father(i)]=get_father(l);
				}
			}
		}
		if (c=='S'){
			scanf("%d%d",&i,&j);
			if (get_father(i)==get_father(j)){
				printf("SUCCESS\n");
			}
			else printf("FAIL\n");
		}
	}
}
示例#3
0
int main()
{
	int t,n,m,c=0,i,j,k,fj,fk;
	char cc;
	scanf("%d",&t);
	scanf("%d%d",&n,&m);
	while(c<t){
#ifdef DBGMSG
		printf("n -- %d\nm -- %d\n",n,m);
#endif
		rank = (int*)malloc(n*sizeof(int));
		father = (int*)malloc(n*sizeof(int));
		ndis = (int*)malloc(n*sizeof(int));
		for(i=0;i<n;i++){
			make_set(i);
		}
		for(i=0;i<m;i++){
			getchar();
			scanf("%c%d%d",&cc,&j,&k);
#ifdef DBGMSG
			printf("#%d : cc -- %c %d\nj -- %d\nk -- %d\n",i,cc,cc,j,k);
#endif
			if((cc != 'A' && cc != 'D') || j > n || k > n){
				fprintf(stderr,"invalid input\n");
				exit(1);
			}
			if(cc == 'A'){
				fj = get_father(j-1);
				fk = get_father(k-1);
				if(fj == fk){
					if(ndis[j-1] == ndis[k-1]){
						printf("In the same gang.\n");
					}else{
						printf("In different gangs.\n");
					}
				}else{
					printf("Not sure yet.\n");
				}
			}else if(cc == 'D'){
				Union(j-1,k-1);
			}
		}
		free(rank);
		free(father);
		free(ndis);
		if(c == (t-1)){
			break;
		}
		scanf("%d%d",&n,&m);
	}
	return 0;
}
示例#4
0
文件: heap.cpp 项目: runys/EDA2
void Heap::shift_up(int pos){
	//Ao chegar na raiz não faz nada
	if(pos==1){
		return;
	}
	//se o elemento for maior que o pai dele, troca de posição com o pai
	else if(this->elements[pos] > this->elements[get_father(pos)]){
		int temp = this->elements[pos];
		this->elements[pos] = this->elements[get_father(pos)];
		this->elements[get_father(pos)] = temp;

		shift_up(get_father(pos));
	}
}
示例#5
0
int get_father(int x){
	if (father[x]==x)
		return x;
	else {
		int tmp=get_father(father[x]);
		father[x]=tmp;
		return tmp;
	}
}
void traverse_up(tree_node *r, int **fdown, int **fup, int level)
{//用先序遍历树的方法遍历每个节点找出其向上的最大权值
 //实际竞赛题目中,使用最简单的数组与下标来构建树,则也可以使用bfs
	if(r != NULL){
		get_father(r, r, fdown, fup, level, 1);
		for(int i = 0; i < r->t_cnt; ++ i)
			traverse_up(r->t_child[i], fdown, fup, level);
	}
}
void decrease_key(position pos,element_type delta,PRIORITY_QUEUE H)
{
    if(pos>H->size)
        return;

    position prev;
    *H->elements[pos]+=delta;
    for(prev=get_father(pos); prev>=1&&((*H->elements[prev])>(*H->elements[pos])); pos=prev,prev=get_father(prev)) {
        swap_elements(prev,pos,H);
    }

    if(((*H->elements[prev])=(*H->elements[pos]))&&((*H->elements[prev])>(*H->elements[pos+1]))) {
        swap_elements(prev,pos+1,H);
    }
}
示例#8
0
static int get_father(int x)
{
	int prev,ndisx = ndis[x];
#ifdef DBGMSG
	printf("current father of %d is %d\n",x+1,father[x]+1);
#endif
	if(father[x] != x){
		prev = father[x];
		father[x] = get_father(father[x]);
		ndis[x] = (ndisx + ndis[prev])%2;
#ifdef DBGMSG
		printf("new father of %d is %d\n",x+1,father[x]+1);
		printf("adjust ndis : ndis[%d] = (ndis[%d] + ndis[%d])%%2 = (%d + %d)%%2 = %d\n",x+1,x+1,prev+1,ndisx,ndis[prev],ndis[x]);
#endif
	}
	return father[x];
}
void get_father(tree_node *r, tree_node *p,
		int **fdown, int **fup, int level, int prev = 1)
{//递归的向上求出r到达 p 祖父节点那一层的最大权值
 //p指代r向上第 prev-1 层的祖父节点
 //prev == 1时,p即为r自己,p->t_fa为r的父节点
 //prev > level时,已经超越了向上的最大层数level
	//递归终止条件
	//若p的父节点为空指针,即p已经为根节点
	if(p->t_fa == NULL)
		return;
	//prev > level是递归结束条件
	if(prev > level)
		return;

	if(prev == level){
		//到达最高的祖父节点时,不访问孩子子树
		fup[r->t_idx][prev] = p->t_fa->t_value;
	}
	else{
		//其他祖父节点可以到达其孩子子树
		for(int i = 0; i < p->t_fa->t_cnt; ++ i)
			//遍历p点的父节点的所有孩子节点,除过自己
			//即遍历p点的所有兄弟节点
			if(p->t_fa->t_child[i] != p){
				//累加父节点的向下权值时要除去p自己这一子树的分支
				//r节点向上prev层的最大权值等于
				//上面prev层的祖父节点的所有孩子节点的向下 level-prev-1 层的最大权值
				//再加上孩子节点自己的权值
				fup[r->t_idx][prev] +=
					fdown[p->t_fa->t_child[i]->t_idx][level - prev - 1];
				fup[r->t_idx][prev] += p->t_fa->t_child[i]->t_value;
			}
		//还要加上p的父节点的权值
		fup[r->t_idx][prev] += p->t_fa->t_value;
	}
	//递归上一层父节点
	get_father(r, p->t_fa, fdown, fup, level, prev + 1);
}