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HDOJ 2121 Ice_cream’s world II 最小樹形圖無根樹

編輯：C++入門知識

HDOJ 2121 Ice_cream’s world II 最小樹形圖無根樹

朱劉算法最小樹形圖無根樹:

建立一個虛擬的根節點,向所有節點連邊,權值為其他所有邊的權值和+1

在求最小樹形圖的時候,記錄和虛擬的根相連的是哪個節點

在這題中,邊是從小往大加的所以直接記錄的是相連的是第幾號邊....

Ice_cream’s world II

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3442 Accepted Submission(s): 823

Problem Description After awarded lands to ACMers, the queen want to choose a city be her capital. This is an important event in ice_cream world, and it also a very difficult problem, because the world have N cities and M roads, every road was directed. Wiskey is a chief engineer in ice_cream world. The queen asked Wiskey must find a suitable location to establish the capital, beautify the roads which let capital can visit each city and the project’s cost as less as better. If Wiskey can’t fulfill the queen’s require, he will be punishing.
Input Every case have two integers N and M (N<=1000, M<=10000), the cities numbered 0…N-1, following M lines, each line contain three integers S, T and C, meaning from S to T have a road will cost C.
Output If no location satisfy the queen’s require, you must be output “impossible”, otherwise, print the minimum cost in this project and suitable city’s number. May be exist many suitable cities, choose the minimum number city. After every case print one blank.

Sample Input

Sample Output

impossible

40 0

Author Wiskey
Source HDU 2007-10 Programming Contest_WarmUp

/* ***********************************************
Author        :CKboss
Created Time  :2015年07月07日 星期二 16時55分48秒
File Name     :HDOJ2121.cpp
************************************************ */

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include

using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn=1100;

int n,m;

struct Edge
{
	int u,v,cost;
}edge[maxn*maxn];

int pre[maxn],id[maxn],vis[maxn];
int in[maxn];
int Root;

int zhuliu(int root,int n,int m,Edge edge[])
{
	int res=0;
	int v;
	while(true)
	{

		for(int i=0;i=2*sum||lens==-1) puts(impossible);
		else printf(%d %d
,lens-sum,Root-m);
		
		putchar(10);
	}
    
    return 0;
}