POJ-2421-Constructing Roads(最小生成树 普利姆)
2016-05-21 19:43
330 查看
D - Constructing Roads
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit
Status
Practice
POJ 2421
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
3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output
179
题意:题意十分简单粗暴,首行给出N,接着是个N*N的矩阵,map[i][j]就代表i到j的权值。接着给出Q,下面Q行,每行两个数字A,B,代表A到B,B到A的权值为0。最后输出最小生成树的权值和就行。
思路:由于是稠密图,所以选用普利姆算法比较合适,大约32ms解决
代码
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit
Status
Practice
POJ 2421
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
3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output
179
题意:题意十分简单粗暴,首行给出N,接着是个N*N的矩阵,map[i][j]就代表i到j的权值。接着给出Q,下面Q行,每行两个数字A,B,代表A到B,B到A的权值为0。最后输出最小生成树的权值和就行。
思路:由于是稠密图,所以选用普利姆算法比较合适,大约32ms解决
代码
#include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> #include<math.h> #include<iomanip> using namespace std; //最小生成树 //普利姆算法 const int maxn=105; const int INF=0x3f3f3f3f; int N;//点的数量 int Q;//已连接的边的数量 int map[maxn][maxn];//矩阵存图 int dis[maxn];//各点到生成树的最小距离 int vis[maxn];//生成树外的点标记为0 void init() { scanf("%d",&N); for(int i=0; i<=N; i++)//初始化图 for(int j=0; j<=N; j++) i==j?map[i][j]=0:map[i][j]=INF; for(int i=1; i<=N; i++)//建图 for(int j=1; j<=N; j++) scanf("%d",&map[i][j]); scanf("%d",&Q); while(Q--)//更新图 { int A,B; scanf("%d%d",&A,&B); map[A][B]=0;//已联通 map[B][A]=0; } for(int i=1; i<=N; i++) dis[i]=map[i][1];//待松弛 memset(vis,0,sizeof(vis)); vis[1]=1;//点1放入生成树 } void Prim() { int min_dis=0; for(int i=1; i<N; i++) { int minn=INF; int point_minn; for(int j=1; j<=N; j++) if(vis[j]==0&&minn>dis[j]) { minn=dis[j]; point_minn=j; } if(minn==INF) break; vis[point_minn]=1; min_dis+=dis[point_minn]; for(int k=1; k<=N; k++) if(vis[k]==0&&dis[k]>map[point_minn][k]) dis[k]=map[point_minn][k]; } printf("%d\n",min_dis); } int main() { init(); Prim(); return 0; }
相关文章推荐
- HDU 5690 查找循环节 数学公式快速幂+乘法逆元(除法取模)
- MySQL数据库解压缩版(免安装版或zip版)无法输入中文,以及与Navicat中文显示一致的问题
- codeforces 670D1 Magic Powder - 1
- 九度OJ 1001:A+B for Matrices
- 在Android开发中使用MVP模式
- 中继器、集线器、网桥、交换机各个组件作用(转))
- 快速排序的分析与实现
- JS学习15(HTML5脚本编程)
- PAT 1015 德才论
- java抽象类练习
- asp.net发布到IIS中出现错误:处理程序“PageHandlerFactory-Integrated”在其模块列表中有一个错误模块“ManagedPipelineHandler”
- 解除百度云浏览器端对下载大文件的限制
- 坐标转换-终结者--OpenCoord提供下载了
- 2016年上半年软考网络工程师考试下午试题参考答案第一时间发布
- mina解决粘包,找不到解码器,数据帧重传的问题
- 16、在JavaScript中,命名的一些规范
- KMP,深入讲解next数组的求解
- Unity发布iOS后 闪屏图片 也就是splash image先黑一下 然后才显示
- 九度OJ 1000:计算A+B
- CRT与Windows的关系【转】