kuangbin 最小生成树
A & M - Jungle Roads HDU - 1301
题意:字母之间的路,求最小生成树
题解:处理好建边以后就是一个Prime
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue>
#include<map>
using namespace std; #define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn = ;
const ll mod = 1e9+;
const double eps = 1e-; bool vis[maxn];
int lowc[maxn]; int Prim(int cost[][maxn],int n)
{
int ans = ;
memset(vis,false,sizeof vis);
vis[] = true;
for(int i=;i<n;i++)
lowc[i] = cost[][i];
for(int i=;i<n;i++)
{
int minc = INF;
int p = -;
for(int j=;j<n;j++)
{
if(!vis[j] && minc > lowc[j])
{
minc = lowc[j];
p = j;
}
}
if(minc == INF)
return -;
ans += minc;
vis[p] = true;
for(int j=;j<n;j++)
if(!vis[j] && lowc[j] > cost[p][j])
lowc[j] = cost[p][j];
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n) && n)
{
int cost[maxn][maxn];
for(int i=;i<maxn;i++)
{
for(int j=;j<maxn;j++)
if(i == j)
cost[i][j] = ;
else
cost[i][j] = INF;
}
for(int i=;i<n;i++)
{
char a;
cin>>a;
int num;
cin>>num;
for(int i=;i<num;i++)
{
char ch;
int len;
cin>>ch>>len;
cost[a-'A'][ch-'A'] = len;
cost[ch-'A'][a-'A'] = len;
}
} int ans = Prim(cost,n);
cout<<ans<<endl;
} }
B - Networking POJ - 1287
题意:有重复边的跑最小生成树
题解:边取小跑Prime
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue> using namespace std; #define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn = 1e2+;
bool vis[maxn];
int lowc[maxn];
int Prim(int cost[][maxn],int n)
{
int ans = ;
memset(vis,false,sizeof vis);
vis[] = true;
for(int i=;i<=n;i++)
lowc[i] = cost[][i];
for(int i=;i<=n;i++)
{
int minc = INF;
int p = -;
for(int j=;j<=n;j++)
if(!vis[j] && minc > lowc[j])
{
minc = lowc[j];
p = j;
}
if(minc == INF)
return -;
ans += minc;
vis[p] = true;
for(int j=;j<=n;j++)
if(!vis[j] && lowc[j] > cost[p][j])
lowc[j] = cost[p][j];
}
return ans; }
int main()
{
int n,m;
while(scanf("%d",&n) && n)
{
scanf("%d",&m);
int cost[maxn][maxn];
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
if(i == j)
cost[i][j] = ;
else
cost[i][j] = INF;
for(int i=;i<m;i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
cost[a][b] = min(cost[a][b],c);
cost[b][a] = min(cost[b][a],c);
cost[a][b] = cost[b][a] = min(cost[a][b],cost[b][a]);
}
int ans = Prim(cost,n);
printf("%d\n",ans);
}
}
C - Building a Space Station POJ - 2031
题意:三维上的球上建最小生成树
题解:处理边即可
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue> using namespace std; #define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn = 1e2+; struct node
{
double x,y,z,r;
}a[maxn];
double dis(double x1,double y1,double z1,double r1,double x2,double y2,double z2,double r2)
{
double len = (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) + (z1-z2)*(z1-z2);
double ans = sqrt(len) - r1 - r2;
return max(0.0,ans);
}
bool vis[maxn];
double lowc[maxn];
double Prime(double cost[][maxn],int n)
{
double ans = ;
memset(vis,,sizeof vis);
vis[] = true;
for(int i=;i<n;i++)
lowc[i] = cost[][i];
for(int i=;i<n;i++)
{
double minc = 9999999999.9;
int p = -;
for(int j=;j<n;j++)
if(!vis[j] && minc > lowc[j])
{
p = j;
minc = lowc[j];
}
ans += minc;
vis[p] = true;
for(int j=;j<n;j++)
if(!vis[j] && lowc[j] > cost[p][j])
lowc[j] = cost[p][j];
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n) && n)
{
double cost[maxn][maxn];
for(int i=;i<n;i++)
scanf("%lf %lf %lf %lf",&a[i].x,&a[i].y,&a[i].z,&a[i].r);
for(int i=;i<n;i++)
for(int j=;j<n;j++)
cost[i][j] = dis(a[i].x,a[i].y,a[i].z,a[i].r,a[j].x,a[j].y,a[j].z,a[j].r);
double ans = Prime(cost,n);
printf("%.3f\n",ans);
}
}
D - Constructing Roads POJ - 2421
题意:邻接矩阵有些边权值为0,求最小生成树
题解:Prim
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue> using namespace std; #define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn = 1e2+; int vis[maxn];
int lowc[maxn];
int cost[maxn][maxn]; int Prim(int cost[][maxn],int n)
{
int ans = ;
memset(vis,,sizeof vis);
vis[] = true;
for(int i=;i<=n;i++)
lowc[i] = cost[][i];
for(int i=;i<=n;i++)
{
int minc = INF;
int p = -;
for(int j=;j<=n;j++)
if(!vis[j] && minc > lowc[j])
{
minc = lowc[j];
p = j;
}
ans += minc;
vis[p] = true;
for(int j=;j<=n;j++)
if(!vis[j] && lowc[j] > cost[p][j])
lowc[j] = cost[p][j];
}
return ans;
}
int main()
{
int n;
scanf("%d",&n);
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
scanf("%d",&cost[i][j]);
int q;
scanf("%d",&q);
for(int i=;i<q;i++)
{
int a,b;
scanf("%d%d",&a,&b);
cost[a][b] = cost[b][a] = ;
}
int ans = Prim(cost,n);
printf("%d\n",ans);
}
E - QS Network ZOJ - 1586
题意:最小生出树,边的值还需要加上两个顶点的值
题解:只需要将顶点的值加到其权值上即可
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue> using namespace std; #define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn = 1e3+; bool vis[maxn];
int lowc[maxn];
int cost[maxn][maxn];
int a[maxn];
int prim(int cost[][maxn],int n)
{
int ans = ;
memset(vis,false,sizeof vis);
vis[] = true;
for(int i=;i<=n;i++)
lowc[i] = cost[][i];
for(int i=;i<=n;i++)
{
int minc = INF;
int p = -;
for(int j=;j<=n;j++)
{
if(!vis[j] && minc > lowc[j])
{
minc = lowc[j];
p = j;
}
}
ans += minc;
vis[p] = true;
for(int j=;j<=n;j++)
if(!vis[j] && lowc[j] > cost[p][j])
lowc[j] = cost[p][j];
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
int ans = ;
scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
for(int i=;i<=n;i++)
for(int j=;j<=n;j++) {
scanf("%d", &cost[i][j]);
cost[i][j] += a[i] + a[j];
}
ans += prim(cost,n);
printf("%d\n",ans);
}
}
F - Truck History POJ - 1789
题意:两个字符串之间不同的个数为边的权值,跑最小生成树
题解:处理边之后跑最小生成树
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue> using namespace std; #define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn = 2e3+; bool vis[maxn];
int lowc[maxn];
int cost[maxn][maxn];
int a[maxn];
int prim(int cost[][maxn],int n)
{
int ans = ;
memset(vis,false,sizeof vis);
vis[] = true;
for(int i=;i<=n;i++)
lowc[i] = cost[][i];
for(int i=;i<=n;i++)
{
int minc = INF;
int p = -;
for(int j=;j<=n;j++)
{
if(!vis[j] && minc > lowc[j])
{
minc = lowc[j];
p = j;
}
}
ans += minc;
vis[p] = true;
for(int j=;j<=n;j++)
if(!vis[j] && lowc[j] > cost[p][j])
lowc[j] = cost[p][j];
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n) && n)
{
char str[maxn][];
for(int i=;i<=n;i++)
scanf("%s",str[i]);
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
{
int tmp = ;
for(int k=;k<;k++)
if(str[i][k] != str[j][k])
tmp++;
cost[i][j] = cost[j][i] = tmp;
}
int ans = prim(cost,n);
printf("The highest possible quality is 1/%d.\n",ans);
}
}
J - Borg Maze POJ - 3026
题意:迷宫之中的MST
题解:先bfs预处理,之后在进行MST,输入的时候有坑点,wa了好几次
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue> using namespace std; #define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn = ;
int tree[maxn][maxn],id[maxn][maxn],dis[maxn][maxn],n,m,tot;
bool vis[maxn][maxn];
int lowc[maxn],visit[maxn];
char mp[maxn][maxn];
int dx[] = {,,,-};
int dy[] = {,,-,};
struct node
{
int x,y;
};
bool check(int x,int y)
{
if(x >= && x <= n && y >= && y <= m && !vis[x][y] && mp[x][y] != '#')
return true;
return false;
}
void bfs(int x,int y)
{
memset(vis,false,sizeof vis);
memset(dis,,sizeof dis);
queue<node>que;
node temp;
temp.x = x;
temp.y = y;
que.push(temp);
vis[x][y] = true;
while(!que.empty())
{
temp = que.front();
que.pop();
for(int i=;i<;i++)
{
int nx = temp.x + dx[i];
int ny = temp.y + dy[i];
if(check(nx,ny))
{
vis[nx][ny] = true;
dis[nx][ny] = dis[temp.x][temp.y] + ;
if(mp[nx][ny] == 'A' || mp[nx][ny] == 'S')
tree[id[x][y]][id[nx][ny]] = tree[id[nx][ny]][id[x][y]] = dis[nx][ny];
que.push(node{nx,ny});
}
}
}
} int Prime(int cost[][maxn],int n)
{
int ans = ;
memset(visit,false,sizeof visit);
visit[] = true;
for(int i=;i<=n;i++)
lowc[i] = cost[][i];
for(int i=;i<=n;i++)
{
int minc = INF;
int p = -;
for(int j=;j<=n;j++)
if(!visit[j] && minc > lowc[j])
{
minc = lowc[j];
p = j;
}
ans += minc;
visit[p] = true;
for(int j=;j<=n;j++)
if(!visit[j] && lowc[j] > cost[p][j])
lowc[j] = cost[p][j];
}
return ans;
} int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int tot = ;
scanf("%d%d ",&m,&n);
for(int i=;i<=n;i++)
gets(mp[i] + );
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
if(mp[i][j] == 'A' || mp[i][j] == 'S')
id[i][j] = ++tot;
for(int i=;i<=tot;i++)
for(int j=;j<=tot;j++)
if(i == j)
tree[i][j] = ;
else
tree[i][j] = INF;
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
if(mp[i][j] == 'A' || mp[i][j] == 'S')
bfs(i,j);
int ans = Prime(tree,tot);
printf("%d\n",ans);
}
}
L - 还是畅通工程 HDU - 1233
题意:直接就是求最小生成树了
题解:板子
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue>
#include<map>
using namespace std; #define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn = ;
const ll mod = 1e9+;
const double eps = 1e-; int lowc[maxn];
bool vis[maxn]; int Prime(int cost[][maxn],int n)
{
int ans = ;
memset(vis,false,sizeof vis);
vis[] = true;
for(int i=;i<n;i++)
lowc[i] = cost[][i];
for(int i=;i<n;i++)
{
int minc = INF;
int p = -;
for(int j=;j<n;j++)
{
if(!vis[j] && minc > lowc[j])
{
minc = lowc[j];
p = j;
}
}
if(minc == INF)
return -;
ans += minc;
vis[p] = true;
for(int j=;j<n;j++)
if(!vis[j] && lowc[j] > cost[p][j])
lowc[j] = cost[p][j];
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n) && n)
{
int cost[maxn][maxn];
memset(cost,,sizeof cost);
for (int i = ; i < n * (n - ) / ; i++)
{
int a, b, l;
scanf("%d%d%d", &a, &b, &l);
cost[a-][b-] = l;
cost[b-][a-] = l;
} int ans = Prime(cost, n);
printf("%d\n", ans);
}
}
N - 畅通工程再续 HDU - 1875
题意:岛之间造桥,只有长度在10~1000才可以造,问最少的花费,花费是桥的长度*100
题解:INF值的判定,INF是长度小于10或者长度大于1000,其余就是Prime的板子
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue>
#include<map>
using namespace std; #define ll long long
#define llu unsigned long long
#define INF 99999999999.9
#define PI acos(-1.0)
const int maxn = ;
const ll mod = 1e9+;
const double eps = 1e-; double lowc[maxn];
bool vis[maxn];
double dis(int x1,int y1,int x2,int y2)
{
double ans = sqrt((double)(x2-x1)*(x2-x1)*1.0 + (y2-y1)*(y2-y1)*1.0);
//cout<<ans<<endl;
return ans;
}
bool check(double x)
{
//cout<<x<<endl;
if(x>=10.0 && x<=1000.0)
return true;
else
return false;
}
double Prime(double cost[][maxn],int n)
{
double ans = ;
memset(vis,false,sizeof vis);
vis[] = true;
for(int i=;i<n;i++)
lowc[i] = cost[][i];
for(int i=;i<n;i++)
{
double minc = INF;
int p = -;
for(int j=;j<n;j++)
{
if(!vis[j] && minc > lowc[j])
{
minc = lowc[j];
p = j;
}
}
if(minc == INF)
return -;
ans += minc;
vis[p] = true;
for(int j=;j<n;j++)
if(!vis[j] && lowc[j] > cost[p][j])
lowc[j] = cost[p][j];
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int x[maxn];
int y[maxn];
double cost[maxn][maxn];
for(int i=;i<n;i++)
scanf("%d %d",&x[i],&y[i]);
for(int i=;i<n;i++)
{
for(int j=;j<n;j++)
{
if(i == j) {
cost[i][j] = ;
continue;
}
else if(check(dis(x[i],y[i],x[j],y[j]))) {
//cout<<dis(x[i], y[i], x[j], y[j])<<endl;
cost[i][j] = dis(x[i], y[i], x[j], y[j]);
cost[j][i] = dis(x[i], y[i], x[j], y[j]);
}
else if(!check(dis(x[i],y[i],x[j],y[j])))
cost[i][j] = INF;
}
} //cout<<cost[0][1]<<endl;
double ans = Prime(cost,n);
if(ans == -)
puts("oh!");
else
printf("%.1f\n",ans * );
}
}
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