Agri-Net
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 25093 |
|
Accepted: 9868 |
Description
Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
The distance between any two farms will not exceed 100,000.
Input
The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.
Output
For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.
Sample Input
4
0 4 9 21
4 0 8 17
9 8 0 16
21 17 16 0
Sample Output
28
很顯然求這個圖的最小生成樹
然后我就打算寫prim了，寫完了之后覺得沒錯了，但是交了好幾遍都wa了，
真心不知道錯誤在哪里，然后去看discuss，才注意到這句話
Physically, they are limited in length to 80 characters, so some lines continue onto others.
當時看題時候沒看明白，看別人說的，沒行還有多余的數據
我就惡心了，我以前寫pascal還有readln可用，現在C語言有不是很熟，用什么代替呢，想了半天沒想出來
還有人說數據范圍不止這些，不過后來驗證后，數據范圍就是100
我就去找題解了
好多題解我發現都沒注意這一點，然后我就納悶了，有看discuss，又有人說，不用注意這句話，我就納悶了，我找了個題解交了之后居然過了
我就在想，是不是我的prim出錯了呢
結果一不小心瞥見，ans居然和vis一塊定義的，定義成short了，
呃……直接無語了
#include<stdio.h>
#include<string.h>
#include<math.h>
#define MAX 505
int map[MAX][MAX],cost[MAX];
short vis[MAX];
int ans;
int n;
void prim()
{
int i,j,mini,min;
memset(vis,0,sizeof(vis));
vis[1]=1;
ans=0;
for (i=1; i<=n; i++)
cost[i]=0x7fffffff;
for (i=2; i<=n; i++)
cost[i]=map[1][i];
for (i=1; i<=n-1; i++)
{
min=0x7fffffff;
for (j=1; j<=n; j++)
if ((vis[j]==0)&&(cost[j]<min))
{
min=cost[j];
mini=j;
}
vis[mini]=1;
ans=ans+min;
for (j=1; j<=n ; j++ )
if ((vis[j]==0)&&(map[mini][j]>0)&&(map[mini][j]<cost[j]))
cost[j]=map[mini][j];
}
}
void init()
{
int i,j;
memset(map,0,sizeof(map));
for (i=1; i<=n ; i++ )
{
for (j=1; j<=n ; j++ )
{
scanf("%d",&map[i][j]);
}
}
}
int main()
{
;
while (scanf("%d",&n)!=EOF)
{
init();
prim();
printf("%d\n",ans);
}
return 0;
}