HDU3768 Shopping(状态压缩DP+spfa)旅行商问题
来源:互联网 发布:ipadcad cad绘图软件 编辑:程序博客网 时间:2024/05/17 03:34
Shopping
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 577 Accepted Submission(s): 197
Problem Description
You have just moved into a new apartment and have a long list of items you need to buy. Unfortunately, to buy this many items requires going to many different stores. You would like to minimize the amount of driving necessary to buy all the items you need.
Your city is organized as a set of intersections connected by roads. Your house and every store is located at some intersection. Your task is to find the shortest route that begins at your house, visits all the stores that you need to shop at, and returns to your house.
Your city is organized as a set of intersections connected by roads. Your house and every store is located at some intersection. Your task is to find the shortest route that begins at your house, visits all the stores that you need to shop at, and returns to your house.
Input
The first line of input contains a single integer, the number of test cases to follow. Each test case begins with a line containing two integers N and M, the number of intersections and roads in the city, respectively. Each of these integers is between 1 and 100000, inclusive. The intersections are numbered from 0 to N-1. Your house is at the intersection numbered 0. M lines follow, each containing three integers X, Y, and D, indicating that the intersections X and Y are connected by a bidirectional road of length D. The following line contains a single integer S, the number of stores you need to visit, which is between 1 and ten, inclusive. The subsequent S lines each contain one integer indicating the intersection at which each store is located. It is possible to reach all of the stores from your house.
Output
For each test case, output a line containing a single integer, the length of the shortest possible shopping trip from your house, visiting all the stores, and returning to your house.
Sample Input
14 60 1 11 2 12 3 13 0 10 2 51 3 53123
Sample Output
4
Source
University of Waterloo Local Contest 2010.07.10
题意:给出n个点,m条路,s(1<=s<=10)个店位于n个地方,现在要求从0点出发走完所有的店的最小花费。
题意:给出n个点,m条路,s(1<=s<=10)个店位于n个地方,现在要求从0点出发走完所有的店的最小花费。
解题:先求出店与店之间的最短路,包括出发点0,然后就是状态压缩dp。
#include<stdio.h>#include<queue>#include<vector>using namespace std;#define mov(a) (1<<(a))const int N = 100005;const int inf = 99999999;struct EDG{ int v,c;};int dis[N],inq[N],mark[N],n,road[15][15],dp[mov(12)][12];vector<EDG>mapt[N];void spfa(int s){ queue<int>q; for(int i=0;i<=n;i++) dis[i]=inf,inq[i]=0; dis[s]=0; q.push(s); while(!q.empty()) { s=q.front(); q.pop(); inq[s]=0; int k=mapt[s].size(); for(int i=0;i<k;i++) { int v=mapt[s][i].v; if(dis[v]>dis[s]+mapt[s][i].c) { dis[v]=dis[s]+mapt[s][i].c; if(inq[v]==0) inq[v]=1,q.push(v); } } }}int main(){ int t,m,s,a,b,store[15]; EDG ss; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(int i=0;i<n;i++) mapt[i].clear(),mark[i]=-1; while(m--) { scanf("%d%d%d",&a,&b,&ss.c); ss.v=b; mapt[a].push_back(ss); ss.v=a; mapt[b].push_back(ss); } scanf("%d",&s); store[0]=0; mark[0]=0; s++; for(int i=1;i<s;i++) { scanf("%d",&store[i]); mark[store[i]]=i; } for(int i=0;i<s;i++)//求出店与店之间的最短路 { spfa(store[i]); for(int j=0;j<n;j++) if(mark[j]>=0) road[i][mark[j]]=dis[j]; } for(int sta=0;sta<mov(s);sta++) for(int i=0;i<s;i++) dp[sta][i]=inf; dp[1][0]=0; for(int sta=1;sta<mov(s);sta++) for(int i=0;i<s;i++) if(dp[sta][i]!=inf) { for(int j=1;j<s;j++) if((sta&mov(j))==0) { if(dp[sta|mov(j)][j]>dp[sta][i]+road[i][j]) dp[sta|mov(j)][j]=dp[sta][i]+road[i][j]; } } int ans=inf; for(int i=0;i<s;i++) if(dp[mov(s)-1][i]+road[i][0]<ans) ans=dp[mov(s)-1][i]+road[i][0]; printf("%d\n",ans); }}
0 0
- HDU3768 Shopping(状态压缩DP+spfa)旅行商问题
- 旅行商问题 (状态压缩DP)
- Shopping(SPFA+DFS HDU3768)
- 旅行商问题(状态压缩dp)
- 旅行商问题(状态压缩dp)
- 三进制状态压缩DP(旅行商问题TSP)HDU3001
- 旅行商问题(状态压缩的DP)
- HDU3768 Shopping 解题报告【SPFA预处理+dfs】
- HDU 4640 Island and study-sister(状态压缩DP+路径压缩)经典 旅行商问题
- POJ 3311 旅行商问题 状态压缩 dp
- 编程题-旅行商问题-状态压缩DP
- 旅行商问题 —— 状态压缩DP
- 最短路+状态压缩dp(旅行商问题)hdu-4568-Hunter
- 二进制状态压缩dp(旅行商TSP)POJ3311
- hdu4568(spfa,状态压缩dp)
- poj - 1170 - Shopping Offers(状态压缩dp)
- POJ 3311 旅行商问题 状态压缩
- 状态压缩动态规划 -- 旅行商问题
- Boys_and_Girls
- GOF23设计模式之模板方法模式(方法回调)的理解与实现之经典
- java学习笔记——第5天
- JQuery小技巧
- STL 中的 set 使用自定义比较运算符
- HDU3768 Shopping(状态压缩DP+spfa)旅行商问题
- uva 679 Dropping Balls
- 虚拟机下Ubuntu修改分辨率
- exFAT硬盘写保护修复&远程登录提示到期
- 逗号运算符和逗号表达式
- Unity3d 对手柄支持的坑
- github之Bug分支
- OC---初识 OC与C的区别 小结
- 第四章 基于对象的编程风格何谓mutable(可变)和const(不可变)