HDU 1102/POJ 2421 Constructing Roads(MST&Prim优化)
来源:互联网 发布:js 观察者模式 实例 编辑:程序博客网 时间:2024/05/22 10:47
Constructing Roads
http://acm.hdu.edu.cn/showproblem.php?pid=1102
http://poj.org/problem?id=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.
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.
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
30 990 692990 0 179692 179 011 2
Sample Output
179
题意:
裸MST,但有部分通路,这里我们将其处理为距离=0就是~
思路:
说一下怎么把Prim算法复杂度从O(N^3)优化到O(N^2):
用一个数组lowcost记录每个点到其他点的最短边。
一开始只知道A到各点的距离,所以lowcost[i] = dis[0][i]
接下来的n-1次循环分两步:
1. 找到剩下的未访问点中lowcost最小的点k:
for (j = 1; j < n; j++)if (!visited[j] && lowcost[j]<min){min = lowcost[j];k = j;}ans += min;visited[k] = true;
2. 通过点k更新所有lowcost:
for (j = 1; j < n; j++)if (!visited[j] && dis[k][j]<lowcost[j])lowcost[j] = dis[k][j];
完整代码:
/*HDU: 0ms,268KB*//*POJ: 0ms,184KB*/#include<cstdio>#include<cstring>using namespace std;const int maxn=101;const int inf = 100001;//okint n;int dis[maxn][maxn];int lowcost[maxn];bool visited[maxn];int prim(){memset(visited, 0, sizeof(visited));int i, j, min, k = 0, ans = 0;for (i = 1; i < n; i++)lowcost[i] = dis[0][i];visited[0] = true;for (i = 1; i < n; i++){min = inf;for (j = 1; j < n; j++)if (!visited[j] && lowcost[j]<min){min = lowcost[j];k = j;//printf("%d ",min);}//else // printf(" ");ans += min;visited[k] = true;for (j = 1; j < n; j++)if (!visited[j] && dis[k][j]<lowcost[j])lowcost[j] = dis[k][j];//printf("\n");}return ans;}int main(){int i, j;while (~scanf("%d", &n)){for (i = 0; i < n; i++)for (j = 0; j < n; j++)scanf("%d", &dis[i][j]);int m, a, b;scanf("%d", &m);while(m--){scanf("%d%d", &a, &b);dis[a - 1][b - 1] = dis[b - 1][a - 1] = 0;//已修路,要修的距离为0}printf("%d\n", prim());}return 0;}
- HDU 1102/POJ 2421 Constructing Roads(MST&Prim优化)
- HDU 1102 && POJ 2421 Constructing Roads (经典MST~Prim)
- [MST]POJ 2421&HDU 1102 Constructing Roads
- poj 2421 Constructing Roads MST
- HDU 1102 Constructing Roads(prim)
- POJ 2421 Constructing Roads (MST)
- poj 2421 hdu 1102 Constructing Roads
- hdu 1102 & poj 2421 Constructing Roads
- poj 2421 Constructing Roads【kruskal & prim】
- Prim 算法, hdu 1102 Constructing Roads
- hdu 1102 Constructing Roads (prim 、kruskal)
- HDU-#1102 Constructing Roads(Prim & Kruskal)
- HDU 1102:Constructing Roads【Kruskal & Prim】
- hdu 1102 Constructing Roads(kruskal || prim)
- hdu 1102 Constructing Roads(prim)
- hdu 1102 Constructing Roads(Prim)
- HDU 1102 Constructing Roads(Prim算法)
- HDU-1102 Constructing Roads(prim)
- 网上常用免费WebServices集合
- CentOS 6下安装nginx
- extjs4学习笔记
- linux驱动入门之LCD驱动
- 卸载VS2010助手
- HDU 1102/POJ 2421 Constructing Roads(MST&Prim优化)
- iOS 滤镜 和 iOS6 中的Core Image技术
- 一步一步学数据结构之1--1(队列--两个栈实现)
- mysql 的启动与关闭
- php header函数的详细解析(转)
- iOS synchronized的作用
- Objective-C 关于空值nil和Nil和NULL和NSNull
- 五大机器视觉测量应用,彰显远心镜头技术优势
- django开发问题杂记