Step6.1.1 1102Constructing Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12150 Accepted Submission(s): 4613
Problem Description
There are N villages, which are numberedfrom 1 to N, and you should build some roads such that every two villages canconnect to each other. We say two village A and B are connected, if and only ifthere is a road between A and B, or there exists a village C such that there isa road between A and C, and C and B are connected.
We know that there are already some roadsbetween some villages and your job is the build some roads such that all thevillages are connect and the length of all the roads built is minimum.
Input
The first line is an integer N (3 <= N<= 100), which is the number of villages. Then come N lines, the i-th ofwhich contains N integers, and the j-th of these N integers is the distance(the distance should be an integer within [1, 1000]) between village i and villagej.
Then there is an integer Q (0 <= Q <=N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1<= a < b <= N), which means the road between village a and village bhas been built.
Output
You should output a line contains aninteger, which is the length of all the roads to be built such that all thevillages are connected, and this value is minimum.
Sample Input
3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output
179
Source
Kicc
題解：
這道題有一個很神奇的題意，就是給你修路的各個城市的花費，但有些路已經修好了，求剩下修的路讓所有的城市連接起來的最小花費。用克魯斯卡爾求最小生成樹，就可以了，先修的路不管花費如何，都標記該這兩個城市之間有路，用克魯斯卡爾算法能計算出有這些城市之間有多少是已經連接到一起了。然後是記錄那些沒修過的路，排序後繼續生成最小生成樹。
源代碼：
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 999999;
typedef struct road
{
intc1,c2,value;
}Road;
int n;
Road rd[10050];
int a[105][105];
int node[105];
void init()
{
memset(a,0,sizeof(a));
memset(node,-1,sizeof(node));
}
bool mycmp(const Road &x,const Road &y)
{
if(x.value< y.value)
return1;
else
return0;
}
int find_set(int x)
{
if(node[x]== -1)
returnx;
returnnode[x] = find_set(node[x]);
}
bool merge(int s1,int s2)
{
intr1 = find_set(s1);
intr2 = find_set(s2);
if(r1== r2)
return0;
if(r1< r2)
node[r2] = r1;
else
node[r1] = r2;
return1;
}
int main()
{
while(cin>> n)
{
init();
for(int i = 1;i <= n;i++)
for(int k = 1;k <= n;k++)
{
cin >> a[i][k];
}
intm;
cin >> m;
intcount = 0;
for(int i = 1;i <= m;i++)
{
intp1,p2;
cin >> p1 >> p2;
if(merge(p1,p2))
count++;
a[p1][p2] = 0;
a[p2][p1] = 0;
}
intnum = 0;
for(int i = 1;i<= n;i++)
{
for(int j = i;j <= n;j++)
{
if(a[i][j])
{
rd[num].c1 = i;
rd[num].c2 = j;
rd[num++].value =a[i][j];
}
}
}
sort(rd,rd+num,mycmp);
intsum = 0;
for(int i = 0;i < num;i++)
{
if(count== n-1)
break;
if(merge(rd[i].c1,rd[i].c2))
{
sum+=rd[i].value;
count++;
}
}
cout<< sum << endl;
}
}