[HDU] 3001 Travelling [状压DP]
来源:互联网 发布:vue.js 2.0视频 百度云 编辑:程序博客网 时间:2024/05/22 14:48
Problem Description
After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can bring him to any city at first but only once.), and of course there are m roads here,following a fee as usual.But Mr Acmer gets bored so easily that he doesn’t want to visit a city more than twice!And he is so mean that he wants to minimize the total fee!He is lazy you see.So he turns to you for help.
Input
There are several test cases,the first line is two intergers n(1<=n<=10) and m,which means he needs to visit n cities and there are m roads he can choose,then m lines follow,each line will include three intergers a,b and c(1<=a,b<=n),means there is a road between a and b and the cost is of course c.Input to the End Of File.
Output
Output the minimum fee that he should pay,or -1 if he can’t find such a route.
Sample Input
2 1
1 2 100
3 2
1 2 40
2 3 50
3 3
1 2 3
1 3 4
2 3 10
Sample Output
100
90
7
http://acm.hdu.edu.cn/showproblem.php?pid=3001
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int MAXN = 95050;int tri[12];int dig[MAXN][11];int edge[11][11], dp[MAXN][11], n, m;void init() { tri[1] = 1; for(int i = 2; i < 12; ++i) tri[i] = 3 * tri[i - 1]; for(int i = 0; i < MAXN; ++i) { int t = i; for(int j = 1; j <= 10; ++j) { dig[i][j] = t % 3; t /= 3; if(t == 0) break; } }}int main(){ init(); while(scanf("%d%d", &n, &m) != EOF) { memset(edge, 0x3f, sizeof edge); memset(dp , 0x3f, sizeof dp ); while(m--) { int x, y, c; scanf("%d%d%d", &x, &y, &c); if(c < edge[x][y]) edge[x][y] = edge[y][x] = c; } int ans = INF; for(int i = 1; i <= n; ++i) dp[tri[i]][i] = 0; for(int s = 0; s < tri[n + 1]; ++s) { int vis_all = 1; for(int i = 1; i <= n; ++i) { if(dig[s][i] == 0) vis_all = 0; if(dp[s][i] == INF) continue; for(int j = 1; j <= n; ++j) { if(edge[i][j] == INF || dig[s][j] == 2) continue; int ns = s + tri[j]; dp[ns][j] = min(dp[ns][j], dp[s][i] + edge[i][j]); } } if(vis_all) { for(int j = 1; j <= n; ++j) { ans = min(ans, dp[s][j]); } } } printf("%d\n", ans == INF ? -1 : ans); } return 0;}
- Travelling - HDU 3001 状压dp
- HDU 3001 Travelling 状压DP
- 状压DP hdu 3001 Travelling
- hdu 3001 Travelling 状压dp
- [HDU] 3001 Travelling [状压DP]
- HDU 3001Travelling (状压DP)
- HDU 3001 Travelling(状压DP)
- HDU 3001 Travelling(状压dp)
- HDU 3001 Travelling(状压dp)
- HDU 3001 Travelling(状压DP)
- hdu 3001 Travelling 状压dp TSP变形
- HDU 3001 Travelling (状压dp三进制)
- HDU 3001 Travelling (状压DP)
- 状压DP——Travelling ( HDU 3001 )
- hdu 3001 Travelling (TSP问题,状压dp)
- hdu 3001 Travelling 三进制状压Dp
- HDU 3001 Travelling 三进制状压DP
- HDU 3001 Travelling 三进制状压DP
- Akka.net分布式数据传输
- 集合(1)
- Ionic2学习基础之Input组件
- halcon字体显示
- 一个软件大学生的心里话
- [HDU] 3001 Travelling [状压DP]
- 50个简单的Jquery代码
- Excel地址转换
- Ubuntu16.04安装WPS
- Spring-AOP @AspectJ切点函数导读
- 306. Additive Number 自己写的,居然ac了,写一遍比看100遍强,看别人对边界的处理纯属浪费时间,不如自己动手,自己手写才能提高
- php学习之旅-3:变量
- Java经典算法题(三)
- Unity-按路径点自动巡航