您的位置:首页 > 其它

POJ 1611 The Suspects 并查集

2016-10-19 22:01 344 查看
Problem:

给了一些队伍,求出0号学生所在队伍的人数,每一个学生可以属于多个团队。

Solution:

并查集的模板题,把各个学生merge,然后直接query就可以了。

#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<vector>
#include<fstream>
#include<list>
#include<iomanip>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
#define ms(s) memset(s,0,sizeof(s))

const double PI = 3.141592653589;
const int INF = 0x3fffffff;

template <typename Type>
class DisjointSet {
struct node {
int parent_;
Type relation_;
node() {
relation_ = 0;
}
};

vector<node> nodes;
vector<int> parents;

public:
Type number_relations_;
bool is_relation;

DisjointSet(int number_nodes) {
parents.resize(number_nodes+1);
is_relation = false;

for(int i = 0; i <= number_nodes; ++i)
parents[i] = i;
}

DisjointSet(int number_nodes, Type number_relations) {
number_relations_ = number_relations;
nodes.resize(number_nodes+1);
is_relation = true;

for(int i = 0; i <= number_nodes; ++i)
nodes[i].parent_ = i;
}

int find(int x) {
if(is_relation == true) {
if(nodes[x].parent_ != x) {
int tdad = nodes[x].parent_; // x's old parent
nodes[x].parent_ = find(nodes[x].parent_);
nodes[x].relation_ = (nodes[x].relation_ + nodes[tdad].relation_) % number_relations_;
}
return nodes[x].parent_;
} else {
if(parents[x] != x)
parents[x] = find(parents[x]);
return parents[x];
}
}

Type get_relation(int x) {
return nodes[x].relation_;
}

void merge(int x, int y) {
parents[find(y)] = find(x);
}

void merge(int x, int y, Type relation) {
// relation: x->y
int ydad = find(y);
nodes[ydad].parent_ = find(x);
nodes[ydad].relation_ = (nodes[x].relation_ + relation + (number_relations_-nodes[y].relation_)) % number_relations_;
}
};

int main() {
//    freopen("/Users/really/Documents/code/input","r",stdin);

int m, n, t, f, num, ans;
while((~scanf("%d%d", &n, &m)) && n){
ans = 1;
DisjointSet<int> ds(n);
for(int i = 0; i < m; ++i){
scanf("%d",&t);
if(t != 0)
scanf("%d",&f);
for(int j = 1 ; j < t; ++j){
scanf("%d", & num);
ds.merge(f, num);
}
}
for(int i = 1; i < n; ++i){
if(ds.find(0) == ds.find(i))
ans++;
}
printf("%d\n",ans);
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: