hdu 3074 Zjnu Stadium (带权并查集)
Zjnu Stadium
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1744 Accepted Submission(s): 660
These days, Busoniya want to hold a large-scale theatrical performance in this stadium. There will be N people go there numbered 1--N. Busoniya has Reserved several seats. To make it funny, he makes M requests for these seats: A B X, which means people numbered B must seat clockwise X distance from people numbered A. For example: A is in column 4th and X is 2, then B must in column 6th (6=4+2).
Now your task is to judge weather the request is correct or not. The rule of your judgement is easy: when a new request has conflicts against the foregoing ones then we define it as incorrect, otherwise it is correct. Please find out all the incorrect requests and count them as R.
For every case:
The first line has two integer N(1<=N<=50,000), M(0<=M<=100,000),separated by a space.
Then M lines follow, each line has 3 integer A(1<=A<=N), B(1<=B<=N), X(0<=X<300) (A!=B), separated by a space.
Output R, represents the number of incorrect request.
1 2 150
3 4 200
1 5 270
2 6 200
6 5 80
4 7 150
8 9 100
4 8 50
1 7 100
9 2 100
Hint:
(PS: the 5th and 10th requests are incorrect)
提议很简单,就是问:
a和b是否在同一个集合,不在就将他们加入到集合,在就对他们进行路径比较,如果给定的路程和实际路程不一样,那就表示冲突,记录下来
输出最后冲突的数量。(注意的是,b在a的前面,也就是b到根的路程要大于a到根的路程)
---------》带权值的路径......
代码:
#include<stdio.h>
#include<stdlib.h>
#define maxn 50005 int father[maxn];
int dist[maxn];
int n,m,re; void init(){
re=;
for(int i=;i<=n;i++){
father[i]=i;
dist[i]=;
}
} int fin(int x){
if(x==father[x]) return x;
int pre=father[x];
father[x]=fin(father[x]);
dist[x]+=dist[pre]; //统计每一个点到根点的距离
return father[x];
} void Union(int x , int y , int val ){
int a=fin(x);
int b=fin(y);
if(a==b){
//注意方向-------顺时针
if(dist[y]-dist[x]!=val)
re++;
}
else{
father[b]=a;
dist[b]=dist[x]-dist[y]+val;
}
} int main(){
int a,b,c;
while(scanf("%d%d",&n,&m)!=EOF){
init();
while(m--){
scanf("%d%d%d",&a,&b,&c);
Union(a,b,c);
}
printf("%d\n",re);
}
return ;
}
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