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#include//判断是否有环，判断是否是一部分，判断是否有点#include#define N 100000int pre[N+10],dis[N+10];int find(int n) {if(pre[n]==n)   return n; return pre[n]=find(pre[n]);}int main() {int min,max,a,b,i

#include#include#define N  200100struct node {int x,count;}pre[N];int find(int n) {if(n!=pre[n].x) {int h=pre[n].x;pre[n].x=find(pre[n].x);pre[n].count=pre[n].count+pre[h].count;

#include#include#define N 110000struct node {    int father,count;}pre[N];int find(int x) {if(x==pre[x].father) {pre[x].count=0;return x;}        int h=pre[x].father;       pre

#include//从大到小不断枚举边直到找到s-t的路径，判断从s可以到t可以用并查集来判断#include//枚举最大的一条边肯定要找和他的值最接近的边，所以要排序#define N 5100#define  inf  0x3fffffffstruct node {int u,v,speed;}map[N];int gcd(int a,int b) {if(b=

#include#include#define N 300int pre[N];int find(int u) {if(u!=pre[u])    pre[u]=find(pre[u]);return pre[u];}struct node {int u,v,speed;}ma[1100];int cmp(const void *a,const vo

#include#include#define N 1100struct node {int u,v,w;}bian[110000];int pre[N];int cmp(const void *a,const void *b) {return (*(struct node *)b).w-(*(struct node *)a).w;}int find(int

#include#include#include#include#define N  700#define NN 2000010int prime[NN];int a[N];struct node {int x,y,w;}ma[N*N];void ispri() {  int i,j;  memset(prime,-1,sizeof(prime)

#include#include#define inf 0x3fffffff#define N 600struct node {int x,y,w;}ma[N*N];int cmp(const void *a,const void *b) {return (*(struct node *)a).w-(*(struct node *)b).w;}int pre[N];int

#include#define N 31000int pre[N];int find(int x) {if(x!=pre[x])    pre[x]=find(pre[x]);return pre[x];}int main() {  int n,m,i,k,s,t;  while(scanf("%d%d",&n,&m),n||m){    for(i=0

#include#define N 1100000struct node {int x,y;}f[N],pre[N];int find(int x) {if(x!=pre[x].x) { int h=pre[x].x; pre[x].x=find(h); pre[x].y=(pre[x].y+pre[h].y)%2;}return pre[x].x;}

//#include#include#define N 110000int pre[N];struct node {int x,y,w;}ma[N];int findd(int x) {if(x!=pre[x]) pre[x]=findd(pre[x]);return pre[x];}int cmp(const void *a,const void*b) {re

/*只需判断等效边和必选边的个数和n-1的关系即可*/#include#include#define N 110struct node { int u,v,w;}f[N*N*2];int cmp(const void *a,const void*b) {return (*(struct node *)a).w-(*(struct node *)b).w;}int pre[

#include#include#define N 550int map[N][N];int f[N][N][5],h[N][5];int pre[N];int find(int n) {if(n!=pre[n])pre[n]=find(pre[n]);return pre[n];}int main() {int n,m,i,j,ans,f1,f2;

http://blog.sina.com.cn/s/blog_803d08c00100y2yy.html#include#include#includeusing namespace std;const int N =10010;struct node {int count,x;}pre[N];int find(int n) {if(n!=pre[n].x)

//不容易啊，终于自己a了一道这种类型的题99M 1 2M 2 4M 4 6M 3 5M 3 7M 1 7//   ////////只要变成这种形式然后路径压缩就行了//#include#includeusing namespace std;const int N=30010;struct node {

#include#include#include#define N 10010#define inf 999999999int pre[N];struct node { int x,deadline;}map[N];int cmp(const void *a,const void *b) {  return (*(struct node *)b).x -(*

#include#define N 1000int pre[N];int find(int n ){return pre[n]=n==pre[n]?n:find(pre[n]);}int main() {int n,m,a,b,i,cnt,f1,f2;while(scanf("%d",&n),n) {scanf("%d",&m);for(i=1;ipre

#include//判断是否有环，判断是否有点，判断是否是一个父节点#include#define N 1000000int pre[N+10],dis[N+10];int find(int n ) {if(pre[n]!=n)return pre[n]=find(pre[n]);return pre[n];}int main() {int n,m,i,a,

#includeint pre[1100];int find(int n) {    if(pre[n]!=n)        return pre[n]=find(pre[n]);    return pre[n];}int main(){    int i,j,a,b,sum,f1,f2,n,m,t,h;    scanf("%d",&t);    wh

#include#include//判断是否有环，判断是否是一个根节点。判断空树的情况#define N 1000000int pre[N+10],dis[N+10],degree[N+10];int find(int n) {return pre[n]=n==pre[n]?n:find(pre[n]);}int main() {int a,b,cnt,flag,f

#include#include#define  N 10000000int pre[N+10],h[N+10],dis[N+10];int find(int n ){return pre[n]=n==pre[n]?n:find(pre[n]);}int main() {int sum,cnt,a,b,f1,f2,n,i,max,min;while(scanf(

#include#include#include#define N 1100struct node { int x,y;}map[N];int s[N];int pre[N];int find(int n) { if(n!=pre[n]) pre[n]=find(pre[n]); return pre[n];}int main() { int n,m,i,

#include#define N 10001#define in  100001int a[in],pre[N],u[in],v[in];int find(int n) { if(n==pre[n])  return n; pre[n]=find(pre[n]);return pre[n];}int main() { int n,m,i,j,x,y,k

dijkstra借鉴http://blog.youkuaiyun.com/wangjian8006/article/details/7871812#include#include#define N  300#define inf 999999999double map[N][N],dis[N];double max(double a,double b) {return a>b?

#include#include#define  N   50010int father[N],link[N];int find(int x) { if(x!=father[x]) {  int h=father[x];  father[x]=find(father[x]);  link[x]=(link[x]+link[h])%3; } return fa

#include#include#define N 100010int father[N],link[N];int find(int x) { if(x!=father[x]) {  int h=father[x];  father[x]=find(father[x]);  link[x]=(link[x]+link[h])%2; } return fath

#include//典型题#include#define N 40010struct node {int x,y,z;}pre[N];struct fff{int a,b,dis;char s[2];}map[N];//储存int find(int x) {if(x!=pre[x].z)  {int h=pre[x].z;pre[x].z=fin

#include#includeusing namespace std;const int N=40010;int pre[N];//并查集int visit[N];//是否经过int ancestor[N];//祖先int dis[N];//储存距离int result[N];//储存结果struct tre{ int x,length;};vec

#include#include#include#includeusing namespace std;const int N=11000;vectorlink[N];//容器来是模拟邻接表int pre[N],a[N],b[N],indegree[N],n,m,sum;char c[N];int find(int x) {//路径压缩if(x!=pre[x