POJ1861 Network(最小生成树)
题意:
有n个顶点,m条边,每条边双向的,并且有一定的长度,现在要求使每个顶点连通起来,并且使总长度最短输出最大边,和边的总数,还有是这些边的情况输出
要点:
这破题,special judge给的输出例子是错的,导致我先用prim算法搞了一个小时还是WA,怒换Kruskal算法10分钟AC了。不过用这题也算理解了special judge的意思,就是有很多种正确输出,只要选一种输出就行。后来发现我的prim算法模板有点问题,要先将map赋予一个比较大的值,否则比较最小值会出错,而且这题要求双向边,我之前没彻底理解prim。
Kruskal算法:
15346658 | Seasonal | 1861 | Accepted | 360K | 63MS | C++ | 1063B | 2016-04-03 10:04:01 |
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#define maxn 1005
using namespace std;
int p[maxn],a[maxn],b[maxn];
int m, n,num;
struct edge
{
int u, v, len;
}e[maxn*maxn];
bool cmp(edge a, edge b)
{
return a.len < b.len;
}
void init()
{
for (int i = 1; i <= m; i++)
p[i] = i;
}
int find(int x)
{
if (p[x] == x) return x;
return p[x] = find(p[x]);
}
bool merge(int x, int y)
{
x = find(x);
y = find(y);
if (x != y)
{
p[x] = y;
return true;
}
return false;
}
int kruskal()
{
init();
sort(e, e + n, cmp);
int max = -1, edges = 0;
num = 0;
for (int i = 0; i < n; i++)
{
if (merge(e[i].u, e[i].v))
{
if (max < e[i].len)
max = e[i].len;
a[num] = e[i].u; b[num++] = e[i].v;
edges++;
}
if (edges + 1 == m)
return max;
}
return -1;
}
int main()
{
int x, y, temp;
scanf("%d%d", &m, &n);
for (int i = 0; i < n; i++)
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].len);
printf("%d\n", kruskal());
printf("%d\n", num);
for (int i = 0; i < num; i++)
printf("%d %d\n", a[i], b[i]);
return 0;
}
Prim算法:
15346706 | Seasonal | 1861 | Accepted | 4124K | 125MS | C++ | 1060B | 2016-04-03 10:26:43 |
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int map[1005][1005], low[1005], pre[1005];
int m, n, num,max;
bool vis[1005];
int a[1005], b[1005];
void prim()
{
int i, j, min, mark;
num = 0; max = 0;
memset(vis, false, sizeof(vis));
for (i = 2; i <= m; i++)
{
low[i] = map[1][i];
pre[i] = 1;//用pre数组记录前驱
}
vis[1] = true;
for (i = 1; i < m; i++)
{
min = 0xffffff;
for (j = 1; j <= m; j++)
if (!vis[j] && low[j] < min)
{
min = low[j];
mark = j;
}
if (max < min)
max = min;
vis[mark] = true;
a[num] = pre[mark]; b[num++] = mark;
for (j = 1; j <= m; j++)
if (!vis[j] && map[mark][j] < low[j])
{
low[j] = map[mark][j];
pre[j] = mark;//此时前驱更新为mark
}
}
}
int main()
{
int x, y, temp;
scanf("%d%d", &m, &n);
for (int i = 1; i <= m; i++)
for (int j = 1; j <= m; j++)
map[i][j] = 0xfffff;//一开始要先赋予一个较大值
for (int i = 0; i < n; i++)
{
scanf("%d%d%d", &x, &y, &temp);
map[x][y] = temp;
map[y][x] = temp;//要双向边
}
prim();
printf("%d\n%d\n", max,num);
for (int i = 0; i < num; i++)
printf("%d %d\n", a[i], b[i]);
return 0;
}