poj 2421 Constructing Roads(最小生成树)
2016-07-18 16:51
267 查看
Constructing Roads
Description
There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village
C such that there is a road between A and C, and C and B are connected.
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
Input
The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village
i and village j.
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
Output
You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.
Sample Input
Sample Output
题意:给一个初始图,再给 m条边,a,b表明a和b不需要花费时间可以直接到达,问最小生成树是多少
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 150
#define INF 99999999
int ma
;
long long d
,vis
;
int n;
long long prim(int s)
{
for(int i=1; i<=n; i++)
d[i]=i==s?0:ma[s][i];
vis[s]=1;
long long ans=0;
for(int i=1; i<n; i++)
{
int maxn=INF,v;
for(int j=1; j<=n; j++)
if(!vis[j]&&maxn>d[j])
{
maxn=d[j];
v=j;
}
vis[v]=1;
ans+=maxn;
for(int j=1; j<=n; j++)
if(!vis[j]&&ma[v][j]<d[j])
d[j]=ma[v][j];
}
return ans;
}
int main()
{
int m;
int s,e;
while(~scanf("%d",&n))
{
memset(vis,0,sizeof(vis));
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
scanf("%d",&ma[i][j]);
scanf("%d",&m);
while(m--)
{
scanf("%d %d",&s,&e);
ma[s][e]=0;
ma[e][s]=0;
}
long long ans=prim(1);
printf("%lld\n",ans);
}
return 0;
}
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 22124 | Accepted: 9431 |
There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village
C such that there is a road between A and C, and C and B are connected.
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
Input
The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village
i and village j.
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
Output
You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.
Sample Input
3 0 990 692 990 0 179 692 179 0 1 1 2
Sample Output
179
题意:给一个初始图,再给 m条边,a,b表明a和b不需要花费时间可以直接到达,问最小生成树是多少
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 150
#define INF 99999999
int ma
;
long long d
,vis
;
int n;
long long prim(int s)
{
for(int i=1; i<=n; i++)
d[i]=i==s?0:ma[s][i];
vis[s]=1;
long long ans=0;
for(int i=1; i<n; i++)
{
int maxn=INF,v;
for(int j=1; j<=n; j++)
if(!vis[j]&&maxn>d[j])
{
maxn=d[j];
v=j;
}
vis[v]=1;
ans+=maxn;
for(int j=1; j<=n; j++)
if(!vis[j]&&ma[v][j]<d[j])
d[j]=ma[v][j];
}
return ans;
}
int main()
{
int m;
int s,e;
while(~scanf("%d",&n))
{
memset(vis,0,sizeof(vis));
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
scanf("%d",&ma[i][j]);
scanf("%d",&m);
while(m--)
{
scanf("%d %d",&s,&e);
ma[s][e]=0;
ma[e][s]=0;
}
long long ans=prim(1);
printf("%lld\n",ans);
}
return 0;
}
相关文章推荐
- HDU 5536 Chip Factory (字典树——序列中查找最大异或和)
- 多用户开源系统哪个好
- Createjs的控件 BitMap、MovieClip、Sprite
- 绝对路径和相对路径
- Activity启动的过程
- 四种生成和解析XML文档的方法详解(介绍+优缺点比较+示例)
- python中sys.exit() os._exit() exit() quit()的简单使用
- iOS开发极光推送显示 开发证书没有通过验证 是否重新上传证书?解决方法
- eslint — js书写规范
- 《征服 C 指针》摘录1:什么是空指针?区分 NULL、0 和 '\0'
- Java JVM:垃圾回收(GC 在什么时候,对什么东西,做了什么事情)
- 学习笔记-斯坦福iOS7-第十七课:摄像头、Core Motion、Application Stat
- Android 第一个程序
- 图片围绕中心旋转
- MFC GDI+基础
- NYOJ2括号配对问题
- mpi学习日志(1):mpi与python
- 杭电 Problem 1047 Integer Inquiry【大数水题】
- 使用头文件winbase.h的错误
- jQuery的deferred对象详解