暢通工程
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7784 Accepted Submission(s): 2969
行對應村莊間道路的成本,每行給出一對正整數,分別是兩個村莊的編號,以及此兩村莊間道路的成本(也是正整數)。爲簡單起見,村莊從1到M編號。當N爲0時,全部輸入結束,相應的結果不要輸出。
#include<stdio.h>
#include<cstdlib>
using namespace std;
typedef struct country{
int x;
int y;
int weight;
}country;
int father[100];
int n,m;
int compare(const void *a,const void *b){
return (*(country*)a).weight-(*(country*)b).weight;
}
void init(){
for(int i=1;i<100;i++){
father[i]=i;
}
}
int find(int i){
if(i!=father[i]){
find(father[i]);
}
return father[i];
}
void unionfather(int a,int b){
int x=find(a),y=find(b);
if(x<y){
father[y]=x;
}
else{
father[x]=y;
}
}
int lestcost(country *cou){
int cost=0,count=0;
init();
for(int i=0;i<n;i++){
if(find(cou[i].x)!=find(cou[i].y)){
unionfather(cou[i].x,cou[i].y);
cost+=cou[i].weight;
count++;
}
}
if(count!=m-1) cost=-1;
return cost;
}
int main(){
country cou[10000];
int cost;
while(scanf("%d%d",&n,&m),n!=0){
for(int i=0;i<n;i++){
scanf("%d%d%d",&cou[i].x,&cou[i].y,&cou[i].weight);
}
qsort(cou,n,sizeof(country),compare);
cost=lestcost(cou);
if(cost==-1) printf("?\n");
else printf("%d\n",cost);
}
return 0;
}