HDU3311-Dig The Wells
来源:互联网 发布:域名别名查询 编辑:程序博客网 时间:2024/05/22 13:17
Dig The Wells
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1420 Accepted Submission(s): 642
Problem Description
You may all know the famous story “Three monks”. Recently they find some places around their temples can been used to dig some wells. It will help them save a lot of time. But to dig the well or build the road to transport the water will cost money. They do not want to cost too much money. Now they want you to find a cheapest plan.
Input
There are several test cases.
Each test case will starts with three numbers n , m, and p in one line, n stands for the number of monks and m stands for the number of places that can been used, p stands for the number of roads between these places. The places the monks stay is signed from 1 to n then the other m places are signed as n + 1 to n + m. (1 <= n <= 5, 0 <= m <= 1000, 0 <=p <= 5000)
Then n + m numbers followed which stands for the value of digging a well in the ith place.
Then p lines followed. Each line will contains three numbers a, b, and c. means build a road between a and b will cost c.
Each test case will starts with three numbers n , m, and p in one line, n stands for the number of monks and m stands for the number of places that can been used, p stands for the number of roads between these places. The places the monks stay is signed from 1 to n then the other m places are signed as n + 1 to n + m. (1 <= n <= 5, 0 <= m <= 1000, 0 <=p <= 5000)
Then n + m numbers followed which stands for the value of digging a well in the ith place.
Then p lines followed. Each line will contains three numbers a, b, and c. means build a road between a and b will cost c.
Output
For each case, output the minimum result you can get in one line.
Sample Input
3 1 31 2 3 41 4 22 4 23 4 4 4 1 45 5 5 5 11 5 12 5 13 5 14 5 1
Sample Output
65
Author
dandelion
Source
HDOJ Monthly Contest – 2010.02.06
Recommend
wxl
题意:选择一些点使得1~n的点与井直接或间接相连,使得总花费最小。1~n+m号点都可以成为井。每个点成为井的花费不同,且修每条路的花费也不同
解题思路:斯坦纳树,建立一个源点0,连接各点,而与各点之间的路程的花费就是各点成为井的花费
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;struct node{int x, y;}pre,nt1;int n, m, p, u, v, w;int s[1500], nt[100009], e[100009], val[100009], cnt;int status;//表示0~n号节点都被选择时的状态+1 int dis[1500][200], vis[1500][200];//dis[i][j]表示以i节点为根选择点集状态为j时的最小值;vis[i][j]表示i节点为点集j时是否在队列中 queue<node> q;void init(){memset(s, -1, sizeof s);memset(vis, 0, sizeof vis);cnt = 0;status = 1 << (n + 1);for (int i = 0; i <= n + m; i++)for (int j = 0; j < status; j++)dis[i][j] = INF;for (int i = 0; i <= n; i++) dis[i][1 << i] = 0;}void SPFA(){while (!q.empty()){pre = q.front();q.pop();vis[pre.x][pre.y] = 0;for (int i = s[pre.x]; ~i; i = nt[i]){if (dis[pre.x][pre.y] + val[i] < dis[e[i]][pre.y]){dis[e[i]][pre.y] = dis[pre.x][pre.y] + val[i];if (!vis[e[i]][pre.y]){nt1 = { e[i],pre.y };q.push(nt1);vis[e[i]][pre.y] = 1;}}}}}void Steiner_Tree(){for (int i = 0; i < status; i++){for (int j = 0; j <= n + m; j++){for (int k = i; k; k = (k - 1) & i)dis[j][i] = min(dis[j][i], dis[j][k] + dis[j][i - k]);if (dis[j][i] != INF){nt1 = { j,i };q.push(nt1);vis[j][i] = 1;}}SPFA();}}int main(){while (~scanf("%d%d%d", &n, &m, &p)){init();for (int i = 1; i <= m + n; i++){scanf("%d", &w);nt[cnt] = s[0], s[0] = cnt, e[cnt] = i, val[cnt++] = w;nt[cnt] = s[i], s[i] = cnt, e[cnt] = 0, val[cnt++] = w;}for (int i = 0; i < p; i++){scanf("%d%d%d", &u, &v, &w);nt[cnt] = s[u], s[u] = cnt, e[cnt] = v, val[cnt++] = w;nt[cnt] = s[v], s[v] = cnt, e[cnt] = u, val[cnt++] = w;}Steiner_Tree();int ans = INF;for (int j = 0; j <= n + m; j++)ans = min(ans, dis[j][status - 1]);printf("%d\n", ans);}return 0;}
阅读全文
0 0
- HDU3311-Dig The Wells
- hdu3311 - Dig The Wells (斯坦纳树 spfa + DP)
- HDOJ 3311 Dig The Wells
- HDU-3311-Dig The Wells
- Hdu 3311 Dig The Wells (综合_斯坦纳树)
- HDU 3311 Dig The Wells(spfa+压缩DP,5级)
- hdu 3311 Dig The Wells (SteinerTree斯坦纳树)
- 【斯坦纳树】 HDOJ 3311 Dig The Wells
- HDU 3311 Dig The Wells(spfa模板)
- hdu3311Dig The Wells
- Wells, Hitler and the World State
- dig
- dig
- Dig Up System Information Using the Terminal
- HDU3311Dig The Wells(斯坦纳树,spfa+状态压缩DP)可作模板
- hdu3311之状态压缩dp
- Bootstrap学习:Wells
- wells 效果 (Bootstrap)
- js 积累
- <漏洞战争软件漏洞分析精要> 学习笔记
- 《The Practice of Programming》读书笔记(一)
- 687. Longest Univalue Path
- 米老师讲课(一年回顾)
- HDU3311-Dig The Wells
- UVa 1584
- 软件测试模型分类
- jsp/servlet第三章第二节Page指令
- API钩取技术
- 搭建Hadoop2.7.3+Hive2.1.1及MySQL(配置Hive+Hadoop)(二)
- 泛型Dao的简单编写
- Valid Sudoku —Leetcode
- Java Main方法中String[] args 与 String args[]区别