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 * );
}
}

kuangbin 最小生成树的更多相关文章

  1. kuangbin最小生成树专题

    网址:https://vjudge.net/contest/66965#overview 第一题: poj1251 裸最小生成树 #include<iostream> #include&l ...

  2. [kuangbin带你飞]专题六 最小生成树

    学习最小生成树已经有一段时间了 做一些比较简单的题还算得心应手..花了三天的时间做完了kuangbin的专题 写一个题解出来记录一下(虽然几乎都是模板题) 做完的感想:有很多地方都要注意 n == 1 ...

  3. [kuangbin]专题六 最小生成树 题解+总结

    kuangbin专题链接:https://vjudge.net/article/752 kuangbin专题十二 基础DP1 题解+总结:https://www.cnblogs.com/RioTian ...

  4. [ An Ac a Day ^_^ ] [kuangbin带你飞]专题八 生成树 UVA 10600 ACM Contest and Blackout 最小生成树+次小生成树

    题意就是求最小生成树和次小生成树 #include<cstdio> #include<iostream> #include<algorithm> #include& ...

  5. [kuangbin带你飞]专题六 最小生成树 POJ 2421 Constructing Roads

    给一个n个点的完全图 再给你m条道路已经修好 问你还需要修多长的路才能让所有村子互通 将给的m个点的路重新加权值为零的边到边集里 然后求最小生成树 #include<cstdio> #in ...

  6. [ An Ac a Day ^_^ ][kuangbin带你飞]专题六 最小生成树 POJ 2031 Building a Space Station

    最小生成树模板题 注意最后输出用%f (从C99开始%f已经不能用于输出double 即 输入用%lf 输出用%f) #include<cstdio> #include<algori ...

  7. [kuangbin带你飞]专题六 最小生成树 POJ 1287 Networking

    最小生成树模板题 跑一次kruskal就可以了 /* *********************************************** Author :Sun Yuefeng Creat ...

  8. [ An Ac a Day ^_^ ] [kuangbin带你飞]专题六 最小生成树 POJ 1251 Jungle Roads

    题意: 有n个点 每个点上有一些道路 求最小生成树 解释下输入格式 A n v1 w1 v2 w2 A点上有n条边 A到v1权值是w1 A到v2权值是w2 思路: 字符串处理之后跑kruskal求最小 ...

  9. 最小生成树 kuangbin专题最后一个题

    题目链接:https://cn.vjudge.net/contest/66965#problem/N 注释:这道题需要用krustra,用prim的话可能会超时.并且在计算距离的时候要尽量减少步骤,具 ...

随机推荐

  1. HCNA修改OSPF基准带宽

    1.拓扑图 2.R1配置ip开启OSPF The device is running! <Huawei>sysEnter system view, return user view wit ...

  2. March 16 2017 Week 11 Thursday

    Adventure may hurt you, but monotony will kill you. 也许冒险会让你受伤,但一成不变会让你灭亡. The very theme of the univ ...

  3. libxml2用xpath进行查找

    xml文档 <?xml version="1.0" encoding="UTF-8"?> <radios> <radio> ...

  4. 一个SAP开发人员的2017总结

    今年的生活如此丰富多彩,不写一点什么怎么对得起这过去的一年? 就当记流水账,若干年之后回来看一定很有意思. 1月 有幸成为SAP Community上2位来自China的SAP mentor之一: 3 ...

  5. git 解决冲突方法

    转载:http://www.cnlvzi.com/index.php/Index/article/id/119 当共享一个项目后提交冲突时 git push -f 强制推送本地的替换服务端 git f ...

  6. Tinkoff Challenge - Final Round (ABC)

    A题:从两个保安中间那钞票 #include <bits/stdc++.h> using namespace std; int main() { int a,b,c; scanf(&quo ...

  7. Going deeper with convolutions(GoogLeNet、Inception)

    从LeNet-5开始,cnn就有了标准的结构:stacked convolutional layers are followed by one or more fully-connected laye ...

  8. 用keytool制作证书并在tomcat配置https服务(四)

    用keytool制作证书并在tomcat配置https服务(一) 用keytool制作证书并在tomcat配置https服务(二) 用keytool制作证书并在tomcat配置https服务(三) 上 ...

  9. 【洛谷P1272】 重建道路

    重建道路 题目链接 一场可怕的地震后,人们用N个牲口棚(1≤N≤150,编号1..N)重建了农夫John的牧场.由于人们没有时间建设多余的道路,所以现在从一个牲口棚到另一个牲口棚的道路是惟一的.因此, ...

  10. Android学习笔记_67_Android MyCrashHandler 中异常处理 UncaughtExceptionHandler

    1.程序中故意抛出异常: public class ExceptionActivity extends Activity { String str; @Override public void onC ...