hdu 1102 Constructing Roads(最小生成树)
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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1102
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
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 b has been built.
3
0 990 692
990 0 179
692 179 0
1
1 2
179
代码如下：
#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f
#define MAXN 517
//创建m二维数组储存图表，low数组记录每2个点间最小权值，visited数组标记某点是否已访问
int m[MAXN][MAXN], low[MAXN], visited[MAXN];
int n;
int prim( )
{
int i, j;
int pos, minn, result=0;
// memset(visited,0,sizeof(visited));
for(i = 1; i <= n; i++)
visited[i] = 0;
visited[1] = 1;
pos = 1; //从某点开始，分别标记和记录该点
for(i = 1; i <= n; i++) //第一次给low数组赋值
{
if(i != pos)
low[i] = m[pos][i];
else
low[i] = 0;
}
for(i = 1; i <= n; i++) //再运行n-1次
{
minn = INF; //找出最小权值并记录位置
pos = -1;
for(j = 1; j <= n; j++)
{
if(visited[j]==0 && minn>low[j])
{
minn = low[j];
pos = j;
}
}
if(pos == -1)
continue;
result += minn; //最小权值累加
visited[pos] = 1; //标记该点
for(j = 1; j <= n; j++) //更新权值
if(!visited[j] && low[j]>m[pos][j])
low[j] = m[pos][j];
}
return result;
}
int main()
{
int tt;
while(~scanf("%d",&n))
{
memset(m,INF,sizeof(m)); //所有权值初始化为最大
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
scanf("%d",&tt);
if(tt < m[j][i])
m[i][j] = m[j][i] = tt;
}
}
int Q, a, b;
scanf("%d",&Q);
for(int i = 0; i < Q; i++)
{
scanf("%d%d",&a,&b);
m[a][b] = m[b][a] = 0;
}
int ans = prim( );
printf("%d\n",ans);
}
return 0;
}
原文：http://blog.csdn.net/u012860063/article/details/38677057
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