poj1611 The Suspects 并查集_小优化

标签：并查集

题目链接
原文出处
并查集基础
(1)

/*    可以通过压缩路径来优化，在查找到祖先节点之后,    将路径上所有它的子孙节点都直接连接到它上面，这样树的高度就大大降低了。*/#include <stdio.h>int num[30005];  //结点所在集合元素的个数int father[30005];void make_set(int x){  ////初始化集合    father[x] = -1;    num[x] = 1;}int find_set(int x){  ////查找x元素所在集合    int r = x, temp;    while(father[r] != -1)  r = father[r];  //找到根节点    while(x != r){  //压缩路径，将路径上r所有的子孙都直接连接到r上        temp = father[x];        father[x] = r;        x = temp;    }    return  x;}void union_(int x, int y){  //合并x, y所在的集合    x = find_set(x), y = find_set(y);    if(x == y)  return ;  //同一个集合则不需要合并    if(num[x] < num[y]) {  //小集合合并到大集合        father[x] = y;        num[y] += num[x];    }    else{        father[y] = x;        num[x] += num[y];    }}int main(){    int n, m, t, x, y, i, j;    while(scanf("%d %d", &n, &m) && (n || m)){        for(i = 0; i < n; i++)  make_set(i);        for(i = 0; i < m; i++){            scanf("%d", &t);            scanf("%d", &x);            for(j = 1; j < t; j++){                scanf("%d", &y);                union_(x, y);            }        }        printf("%d\n", num[find_set(0)]);    }    return 0;}

(2)

/*    对于根节点来说，father[]本身就可以记录节点数目，只不过值是负的。*/#include <stdio.h>int father[30005];void make_set(int x){  //初始化集合    father[x] = -1;}int find_set(int x){  //查找x元素所在集合    int r = x, temp;    while(father[r] > 0)  r = father[r];  ////找到根节点    while(x != r){  //压缩路径，将路径上r所有的子孙都直接连接到r上        temp = father[x];        father[x] = r;        x = temp;    }    return  x;}void union_(int x, int y){  //合并x, y所在的集合    x = find_set(x), y = find_set(y);    if(x == y)  return ;  //同一个集合则不需要合并    if(father[x] < father[y]) {  ////小集合合并到大集合(负数比较), <        father[x] += father[y];        father[y] = x;        }    else{        father[y] += father[x];        father[x] = y;        }}int main(){    int n, m, t, x, y, i, j;    while(scanf("%d %d", &n, &m) && (n || m)){        for(i = 0; i < n; i++)  make_set(i);        for(i = 0; i < m; i++){            scanf("%d", &t);            scanf("%d", &x);            for(j = 1; j < t; j++){                scanf("%d", &y);                union_(x, y);            }        }        printf("%d\n", -father[find_set(0)]);  //结点数目为负值    }    return 0;}