带权并查集 La 3027

来源：互联网发布：长安城知乎编辑：程序博客网时间：2024/05/23 01:20

题目链接： http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4075

每参考题解的时候其实我已经自己模拟把并查集改为带权并查集了

但是因为取模弄错 WA了几个小时。。。。。。。。。。。。

其实在回溯的时候，修改权值就行，然后查询之前，还需要再维护一次

先贴我的垃圾代码，好TMD冗杂：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <functional>using namespace std;#define MAXN 20010#define MOD 1000int father[MAXN],dis[MAXN];void makeset(int n){    for(int i=0;i<=n;i++)    {        father[i]=i;        dis[i]=0;    }}inline int ABS(int a,int b){    if(a-b>0)return a-b;    else return b-a;}int Find1(int x){    if(x != father[x])    {        int t = Find1(father[x]);        dis[x]+=dis[father[x]];        father[x] =t;        return father[x];    }    else    return x;}void Union(int x, int y){    /*调用Union之前必须先确定x,y不在同一个集合，然后把x的父节点改为y*/    dis[x]=ABS(x,y)%MOD;    int t=Find1(y);    dis[y]+=dis[t];    dis[x]+=dis[y];    y=t;    father[x]=y;}int main(){    //freopen("La 3027.txt","r",stdin);    int n,ncase,u,v;    char c;    scanf("%d",&ncase);    while(ncase--)    {        scanf("%d",&n);        makeset(n);        while(1)        {            while((c=getchar()) != 'O' && c != 'E' && c != 'I');            if(c == 'O')break;            if(c == 'E')            {                scanf("%d",&u);                Find1(u);                printf("%d\n",dis[u]);            }            else            {                if(c == 'I')                {                    scanf("%d%d",&u,&v);                    Union(u,v);                    //dis[u]=(ABS(u,v))%MOD;                    //father[u]=v;                }            }        }    }    return 0;}

然后看了刘汝佳的代码果然大牛，比我的代码短了一半......

// LA3027 Corporative Network// Rujia Liu#include<iostream>#include<string>#include<algorithm>using namespace std;const int maxn = 20000 + 10;int pa[maxn], d[maxn];int findset( int x ) {  if (pa[x] != x) {    int root = findset( pa[x] );    d[x] += d[pa[x]];    return pa[x] = root;   } else return x;}int main() {  int T;  cin >> T;  while(T--) {    int n, u, v;    string cmd;    cin >> n;    for(int i = 1; i <= n; i++) { pa[i] = i; d[i] = 0; }    while(cin >> cmd && cmd[0] != 'O') {      if(cmd[0] == 'E') { cin >> u; findset(u); cout << d[u] << endl; }      if(cmd[0] == 'I') { cin >> u >> v; pa[u] = v; d[u] = abs(u-v) % 1000; }    }  }  return 0;}

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