1)题目中所指的所谓的"数据不足"其实就是指，还有孤立节点(除了根节点)。只要对其进行判断即可

``````for( i = 1 ; i <= n ; ++i){
if(father[i] == i){
count++;
}
}

if(count == 1){
printf("%d\n",sum);
}``````

2)kruscal算法的实现如下:

``````int kruscal(){
int i;
int sum = 0;

//初始化
for( i = 1 ; i<= n ; ++i){
father[i] = i;
}

//合并
for( i = 1 ; i <= n ; ++i){
int fx = find(e[i].begin);
int fy = find(e[i].end);

if(fx != fy){
father[fx] = fy;
sum += e[i].weight;
}
}

return sum;
}``````

``````/*
* 1863_1.cpp
*
*  Created on: 2013年8月26日
*/
#include <iostream>
using namespace std;

struct edge{
int begin;
int end;
int weight;
};

const int maxn = 110;
edge e[maxn*maxn];
int father[maxn];
int n;

int find(int a){
if( a == father[a]){
return a;
}

father[a] = find(father[a]);
return father[a];
}

int kruscal(){
int i;
int sum = 0;

//初始化
for( i = 1 ; i<= n ; ++i){
father[i] = i;
}

//合并
for( i = 1 ; i <= n ; ++i){
int fx = find(e[i].begin);
int fy = find(e[i].end);

if(fx != fy){
father[fx] = fy;
sum += e[i].weight;
}
}

return sum;
}

bool compare(const edge& a , const edge& b){
return a.weight < b.weight;
}
int main(){
int m;
while(scanf("%d%d",&m,&n)!=EOF,m){
int i;
memset(father,0,sizeof(father));

for( i = 1 ; i <= maxn*maxn ; ++i){
e[i].begin = -1;
e[i].end = -1;
e[i].weight = 0;
}

for(i = 1 ; i <= m ; ++i){
scanf("%d%d%d",&e[i].begin,&e[i].end,&e[i].weight);
}

sort(e + 1, e + m + 1 , compare );

int sum = kruscal();
int count = 0;

for( i = 1 ; i <= n ; ++i){
if(father[i] == i){
count++;
}
}

if(count == 1){
printf("%d\n",sum);
}else{
printf("?\n");
}
}
}``````