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 =  50;
const ll mod = 1e9+7;
const double eps = 1e-8;

bool vis[maxn];
int lowc[maxn];

int Prim(int cost[][maxn],int n)
{
    int ans = 0;
    memset(vis,false,sizeof vis);
    vis[0] = true;
    for(int i=1;i<n;i++)
        lowc[i] = cost[0][i];
    for(int i=1;i<n;i++)
    {
        int minc = INF;
        int p = -1;
        for(int j=0;j<n;j++)
        {
            if(!vis[j] && minc > lowc[j])
            {
                minc = lowc[j];
                p = j;
            }
        }
        if(minc == INF)
            return -1;
        ans += minc;
        vis[p] = true;
        for(int j=0;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=0;i<maxn;i++)
        {
            for(int j=0;j<maxn;j++)
                if(i == j)
                    cost[i][j] = 0;
                else
                    cost[i][j] = INF;
        }
        for(int i=1;i<n;i++)
        {
            char a;
            cin>>a;
            int num;
            cin>>num;
            for(int i=0;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;
    }

}

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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 =  1005;
const ll mod = 1e9+7;
const double eps = 1e-8;

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 = 0;
    memset(vis,false,sizeof vis);
    vis[0] = true;
    for(int i=1;i<n;i++)
        lowc[i] = cost[0][i];
    for(int i=1;i<n;i++)
    {
        double minc = INF;
        int p = -1;
        for(int j=0;j<n;j++)
        {
            if(!vis[j] && minc > lowc[j])
            {
                minc = lowc[j];
                p = j;
            }
        }
        if(minc == INF)
            return -1;
        ans += minc;
        vis[p] = true;
        for(int j=0;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=0;i<n;i++)
            scanf("%d %d",&x[i],&y[i]);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(i == j) {
                    cost[i][j] = 0;
                    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 == -1)
            puts("oh!");
        else
            printf("%.1f\n",ans * 100);
    }
}

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