zoukankan      html  css  js  c++  java
  • zoj 1586

    //zoj 1586
    #include<iostream>
    #include<cstdio>
    using namespace std;
    #define N 1005
    int a[N][N],low[N],n,ans;
    int b[N];

    int min(int x,int y)
    { return x<y?x:y; }

    void prim(int u0)
    {
    int i,j,m,k;
    ans=0;
    for (i=1;i<n;i++) low[i]=a[u0][i];
    low[u0]=-1;
    for (i=1;i<n;i++)
    {
    m=1<<20;
    for (j=0;j<n;j++)
    if (low[j]!=-1&&low[j]<m)
    { m=low[j]; k=j; }
    ans+=m;
    low[k]=-1;
    for (j=0;j<n;j++)
    if (low[j]!=-1)low[j]=min(low[j],a[k][j]);
    }
    }

    int main()
    {
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
    scanf("%d",&n);
    for(i=0;i<n;i++) scanf("%d",&b[i]);

    for(i=0;i<n;i++)
    {
    for(j=0;j<n;j++)
    {
    scanf("%d",&a[i][j]);
    a[i][j]+=b[i]+b[j];
    }
    }
    prim(0);
    cout<<ans<<endl;
    }
    return 0;
    }

    ***************************************************************************************************

    #include<iostream>
    #include<cstdio>
    using namespace std;
    #define N 1005
    int a[N][N],low[N],n,ans;
    int b[N];

    int min(int x,int y)
    { return x<y?x:y; }

    void prim(int u0)
    {
    int i,j,m,k;
    ans=0;
    for (i=1;i<n;i++) low[i]=a[u0][i];
    low[u0]=-1;
    for (i=1;i<n;i++)
    {
    m=1<<20;
    for (j=0;j<n;j++)
    if (low[j]!=-1&&low[j]<m)
    { m=low[j]; k=j; }
    ans+=m;
    low[k]=-1;
    for (j=0;j<n;j++)
    if (low[j]!=-1)low[j]=min(low[j],a[k][j]);
    }
    }

    int main()
    {
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
    scanf("%d",&n);
    for(i=0;i<n;i++) scanf("%d",&b[i]);

    for(i=0;i<n;i++)
    {
    for(j=0;j<n;j++)
    {
    scanf("%d",&a[i][j]);
    if(i=j) a[i][j]=1<<20; //对角线
    else a[i][j]+=b[i]+b[j];
    }
    }
    prim(0);
    cout<<ans<<endl;
    }
    return 0;
    }

    **************************************************************************888


    #include<iostream>
    #include<cstdio>
    using namespace std;
    #define N 1005
    #define INF 1000000
    int a[N][N],low[N],n,ans;
    int b[N];

    int min(int x,int y)
    { return x<y?x:y; }

    void prim(int u0)
    {
    int i,j,m,k;
    ans=0;
    for (i=1;i<n;i++) low[i]=a[u0][i];
    low[u0]=-1;
    for (i=1;i<n;i++)
    {
    m=1<<20;
    for (j=0;j<n;j++)
    if (low[j]!=-1&&low[j]<m)
    { m=low[j]; k=j; }
    ans+=m;
    low[k]=-1;
    for (j=0;j<n;j++)
    if (low[j]!=-1)low[j]=min(low[j],a[k][j]);
    }
    }

    int main()
    {
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
    scanf("%d",&n);
    for(i=0;i<n;i++) scanf("%d",&b[i]);

    for(i=0;i<n;i++)
    {
    for(j=0;j<n;j++)
    {
    scanf("%d",&a[i][j]);
    if(i=j) a[i][j]=INF; //对角线
    else a[i][j]+=b[i]+b[j];
    }
    }
    prim(0);
    cout<<ans<<endl;
    }
    return 0;
    }

  • 相关阅读:
    企业如何推行白盒测试
    Java命名规范
    MobileVLC for iphoneos4.3
    用Android NDK编译FFmpeg
    Linux 下Android 开发环境搭建 ---CentOS
    为什么要做白盒测试
    vlc的第三方库contrib的修改之一:live库的修改
    VC命名规范
    POJ 1470 Closest Common Ancestors (LCA入门题)
    HDU 4407 Sum 第37届ACM/ICPC 金华赛区 第1008题 (容斥原理)
  • 原文地址:https://www.cnblogs.com/2014acm/p/3903262.html
Copyright © 2011-2022 走看看