POJ1258
简单Prim 邻接矩阵存边
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int INF=100001;
int n;
int a[200][200],dis[200];
bool vis[200];
int Prim()
{
memset(vis,false,sizeof(vis));
for (int i=1;i<=n;i++) dis[i]=a[1][i];
int ans=0;
vis[1]=true;
for (int i=1;i<=n-1;i++)
{
int tmp=INF,k=0;
for (int j=1;j<=n;j++)
{
if (!vis[j]&&dis[j]<tmp)
{
tmp=dis[j];
k=j;
}
}
if (k==0) return -1;
vis[k]=true;
ans=ans+dis[k];
for (int j=1;j<=n;j++)
{
if (!vis[j]&&dis[j]>a[k][j])
dis[j]=a[k][j];
}
}
return ans;
}
int main()
{
while (scanf("%d",&n)!=EOF)
{
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
scanf("%d",&a[i][j]);
printf("%d\n",Prim());
}
return 0;
}
POJ2485
简单Prim 变形,求最小生成树中最短的边。
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int INF=1<<30;
int dis[505],a[505][505];
bool vis[505];
int n;
int Prim()
{
memset(vis,false,sizeof(vis));
for (int i=1;i<=n;i++) dis[i]=a[1][i];
int ans=0;
vis[1]=true;
for (int i=1;i<=n-1;i++)
{
int tmp=INF,k=0;
for (int j=1;j<=n;j++)
{
if (!vis[j]&&dis[j]<tmp)
{
tmp=dis[j];
k=j;
}
}
if (k==0) return -1;
ans=max(ans,dis[k]);
vis[k]=true;
for (int j=1;j<=n;j++)
{
if (!vis[j]&&dis[j]>a[k][j])
dis[j]=a[k][j];
}
}
return ans;
}
int main()
{
int cases;
scanf("%d",&cases);
while (cases--)
{
scanf("%d",&n);
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
scanf("%d",&a[i][j]);
printf("%d\n",Prim());
}
return 0;
}
简单Prim
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
bool vis[3000];
int dis[3000],a[2500][2500];
int n;
int Prim()
{
memset(vis,false,sizeof(vis));
for (int i=1;i<=n;i++) dis[i]=a[1][i];
int ans=0;vis[1]=true;
for (int i=1;i<=n-1;i++)
{
int tmp=100,k=0;
for (int j=1;j<=n;j++)
{
if (!vis[j]&&dis[j]<tmp)
{
tmp=dis[j];
k=j;
}
}
if (k==0) return -1;
ans+=dis[k];
vis[k]=true;
for (int j=1;j<=n;j++)
{
if (!vis[j]&&dis[j]>a[k][j])
dis[j]=a[k][j];
}
}
return ans;
}
int main()
{
char ch[2500][10];
while (scanf("%d",&n),n!=0)
{
for (int i=1;i<=n;i++)
{
cin>>ch[i];
}
for (int i=1;i<=n;i++)
{
for (int j=i+1;j<=n;j++)
{
int t=0;
for (int k=0;k<7;k++)
if (ch[i][k]!=ch[j][k]) t++;
a[i][j]=a[j][i]=t;
}
}
printf("The highest possible quality is 1/%d.\n",Prim());
}
return 0;
}
POJ1751 ZOJ2048
注意 两个OJ 的输入方式不同= =、
题目大意:在图中,有些边已经连接,输出最小生成树的方案。
分析:连接的边赋值为0即可。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#define Maxn 760
using namespace std;
struct Node{
double x,y;
}b[Maxn];
double dis[Maxn],a[Maxn][Maxn];
bool vis[Maxn];
int pre[Maxn];
int n,m;
double Distance(int i,int j){
double tmp=sqrt((b[i].x-b[j].x)*(b[i].x-b[j].x)+
(b[i].y-b[j].y)*(b[i].y-b[j].y));
return tmp;
}
int Prim(){
for (int i=1;i<=n;i++){
dis[i]=a[1][i];pre[i]=1;
}
memset(vis,false,sizeof(vis));
vis[1]=true;
for (int i=1;i<=n-1;i++){
double Min=100000;int k=0;
for (int j=1;j<=n;j++){
if (!vis[j]&&dis[j]<Min){
Min=dis[j];
k=j;
}
}
if (k==0) break;
vis[k]=true;
if (Min!=0) printf("%d %d\n",pre[k],k);
for (int j=1;j<=n;j++){
if (!vis[j]&&dis[j]>a[k][j]){
dis[j]=a[k][j];
pre[j]=k;
}
}
}
}
int main(){
int cases;
int u,v;
scanf("%d",&cases);
while (cases--){
scanf("%d",&n);
for (int i=1;i<=n;i++){
scanf("%lf%lf",&b[i].x,&b[i].y);
}
for (int i=1;i<=n;i++){
for (int j=i+1;j<=n;j++){
a[i][j]=a[j][i]=Distance(i,j);
}
}
scanf("%d",&m);
for (int i=1;i<=m;i++){
scanf("%d%d",&u,&v);
a[u][v]=a[v][u]=0;
}
Prim();
if (cases>0) putchar('\n');
}
return 0;
}