hdu 3371 Connect the Cities <prim>
来源:互联网 发布:mac版网游有哪些 编辑:程序博客网 时间:2024/05/16 17:09
Connect the Cities
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4938 Accepted Submission(s): 1460
Problem Description
In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.
Input
The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
Output
For each case, output the least money you need to take, if it’s impossible, just output -1.
Sample Input
16 4 31 4 22 6 12 3 53 4 332 1 22 1 33 4 5 6
Sample Output
1
Author
dandelion
Source
HDOJ Monthly Contest – 2010.04.04
Recommend
lcy
城市被淹,给出城市距离,给出哪些城市已经联通,求联通的最小代价 prime
#include<iostream> #include<stdio.h> using namespace std; bool b[501]; int d[501]; int a[501][501]; const int maxt=1000000000; int main() { int m,n,i,j,k,t,p1,p2,min,tree,p,q,c,u,z; scanf("%d",&t); for(;t>=1;t--) { scanf("%d%d%d",&n,&m,&z); for(i=0;i<=n;i++) { for(j=i;j<=n;j++) { a[i][j]=a[j][i]=maxt; } } for(i=0;i<m;i++) { scanf("%d%d%d",&p,&q,&c); if(c<a[p][q])a[p][q]=a[q][p]=c; } for(j=0;j<z;j++) { scanf("%d%d",&k,&p1); for(i=2;i<=k;i++) { scanf("%d",&p2); { a[p1][p2]=a[p2][p1]=0; } } } min=maxt,u=0; tree=0; for(i=1;i<=n;i++) { d[i]=a[1][i]; b[i]=false; } b[1]=true; for(i=1;i<n;i++) { min=maxt; for(j=1;j<=n;j++) { if(b[j]==false&&d[j]<min) { min=d[j]; u=j; } } tree=tree+min; if(tree>=maxt)break; b[u]=true; for(j=1;j<=n;j++) { if(b[j]==false&&a[u][j]<d[j]) { d[j]=a[u][j]; } } } if(tree>=maxt){printf("%d\n",-1);} else{printf("%d\n",tree);} } }
- hdu 3371 Connect the Cities(prim)
- hdu 3371 Connect the Cities <prim>
- hdu 3371 Connect the Cities【kruskal&prim】
- hdu 3371 Connect the Cities (Prim)
- hdu 3371 Connect the cities (prim)
- hdu 3371 Connect the Cities(prim算法)
- HDU 3371 Connect the Cities(Prim,Kruskal)
- hdu 3371 Connect the Cities 最小生成树prim
- Connect the Cities(Prim)
- hdu 3371 Connect the Cities
- Hdu-3371 Connect the Cities
- HDU 3371 Connect the Cities
- HDU-3371-Connect the Cities
- hdu 3371 Connect the Cities
- HDU 3371 Connect the Cities
- Hdu 3371 - Connect the Cities
- hdu 3371 Connect the Cities
- hdu 3371 Connect the Cities
- Android中获取应用系统中应用信息接口
- Linux IPC总结(全)
- CCNP:HSRP介绍及校园网的应用
- scp
- Android创业指导,学习+分享,教你轻松玩转Android Market
- hdu 3371 Connect the Cities <prim>
- java的volatile是什么意思
- visual studio 2012发现的小bug
- http://www.devbean.info/
- Weblogic_Server与EJB
- XML解析技术
- 奥尔良烤翅(自调腌料做法)
- Drupal commerce 性能优化 (Drupal commerce performance optimize)
- hau1198 这个代码一直w 思路对啊 还有特殊情况吗? 求指教