--最小生成树的处理(始终抓取最小边的克鲁斯卡尔算法)
内容导读
互联网集市收集整理的这篇技术教程文章主要介绍了--最小生成树的处理(始终抓取最小边的克鲁斯卡尔算法),小编现在分享给大家,供广大互联网技能从业者学习和参考。文章包含4606字,纯文字阅读大概需要7分钟。
内容图文
![--最小生成树的处理(始终抓取最小边的克鲁斯卡尔算法)](/upload/InfoBanner/zyjiaocheng/1044/3e0a6cd4333242ab9e1019d2a08838fd.jpg)
Consider yourself lucky! Consider yourself lucky to be still breathing and having fun participating in
this contest. But we apprehend that many of your descendants may not have this luxury. For, as you
know, we are the dwellers of one of the most polluted cities on earth. Pollution is everywhere, both in
the environment and in society and our lack of consciousness is simply aggravating the situation.
However, for the time being, we will consider only one type of pollution - the sound pollution. The
loudness or intensity level of sound is usually measured in decibels and sound having intensity level 130
decibels or higher is considered painful. The intensity level of normal conversation is 6065 decibels and
that of heavy tra?c is 7080 decibels.
Consider the following city map where the edges refer to streets and the nodes refer to crossings.
The integer on each edge is the average intensity level of sound (in decibels) in the corresponding street.
To get from crossing A to crossing G you may follow the following path: A-C-F-G. In that case
you must be capable of tolerating sound intensity as high as 140 decibels. For the paths A-B-E-G,
A-B-D-G and A-C-F-D-G you must tolerate respectively 90, 120 and 80 decibels of sound intensity.
There are other paths, too. However, it is clear that A-C-F-D-G is the most comfortable path since
it does not demand you to tolerate more than 80 decibels.
In this problem, given a city map you are required to determine the minimum sound intensity level
you must be able to tolerate in order to get from a given crossing to another.
Input
The input may contain multiple test cases.
The rst line of each test case contains three integers C( 100), S( 1000) and Q( 10000) where
C indicates the number of crossings (crossings are numbered using distinct integers ranging from 1 to
C), S represents the number of streets and Q is the number of queries.
Each of the next S lines contains three integers: c1; c2 and d indicating that the average sound
intensity level on the street connecting the crossings c1 and c2 (c1 ?= c2) is d decibels.
Each of the next Q lines contains two integers c1 and c2 (c1 ?= c2) asking for the minimum sound
intensity level you must be able to tolerate in order to get from crossing c1 to crossing c2.
The input will terminate with three zeros form C, S and Q.
Output
For each test case in the input rst output the test case number (starting from 1) as shown in the
sample output. Then for each query in the input print a line giving the minimum sound intensity level
(in decibels) you must be able to tolerate in order to get from the rst to the second crossing in the
query. If there exists no path between them just print the line \no path".
Print a blank line between two consecutive test cases.
Sample Input
7 9 3
1 2 50
1 3 60
2 4 120
2 5 90
3 6 50
4 6 80
4 7 70
5 7 40
6 7 140
1 7
2 6
6 2
7 6 3
1 2 50
1 3 60
2 4 120
3 6 50
4 6 80
5 7 40
7 5
1 7
2 4
0 0 0
Sample Output
Case #1
80
60
60
Case #2
40
no path
80
#include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; int pre[110]; struct node{ int x,y,cost; }edges[10010]; bool cmp(node a,node b){ return a.cost < b.cost; } void init(){ for(int i = 0;i<=110;i++){ pre[i] = i; } } int finds(int x){ int t = x; while(x != pre[x]){ x = pre[x]; } while(t != pre[t]){ int m = t; t = pre[t]; pre[m] = x; } return x; } int main(){ int cases = 1; int flag = 1; int n,m,q; int ms[110][110]; while(scanf("%d%d%d",&n,&m,&q),n||m||q){ memset(ms,0,sizeof(ms)); int x,y,c; init(); for(int i = 0;i<m;i++){ scanf("%d%d%d",&x,&y,&c); edges[i].x = x; edges[i].y = y; edges[i].cost = c; } sort(edges,edges+m,cmp); int qs_u[10010],qs_v[10010]; for(int i = 0;i<q;i++){ scanf("%d%d",&qs_u[i],&qs_v[i]); } for(int i=0 ;i<m;i++){ int u = finds(edges[i].x); int v = finds(edges[i].y); if(u != v){ pre[u] = v; ms[edges[i].x][edges[i].y] = ms[edges[i].y][edges[i].x] = edges[i].cost; } } if(flag ==1 )flag = 0; else puts(""); printf("Case #%d\n",cases++); for(int i = 0;i<q;i++){ int is_find = 0; int vis[110]; int s = qs_u[i]; int e = qs_v[i]; queue<int> qq; queue<int> cc; memset(vis,0,sizeof(vis)); qq.push(s); vis[s] = 1; cc.push(0); while(!qq.empty()){ int z = qq.front(); int cf = cc.front(); qq.pop(); cc.pop(); if(z == e){ is_find = 1; printf("%d\n",cf); break; } for(int j = 1;j<=n;j++){ if(ms[z][j]&& vis[j] == 0){ vis[j] = 1; qq.push(j); cc.push(max(cf,ms[z][j])); } } } if(is_find == 0)puts("no path"); } } }
原文:http://www.cnblogs.com/lovelystone/p/4729370.html
内容总结
以上是互联网集市为您收集整理的--最小生成树的处理(始终抓取最小边的克鲁斯卡尔算法)全部内容,希望文章能够帮你解决--最小生成树的处理(始终抓取最小边的克鲁斯卡尔算法)所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错,欢迎将互联网集市网站推荐给程序员好友。
内容备注
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 gblab@vip.qq.com 举报,一经查实,本站将立刻删除。
内容手机端
扫描二维码推送至手机访问。