HDU

HDU - 2121 ：http://acm.hdu.edu.cn/showproblem.php?pid=2121

比较好的朱刘算法blog：https://blog.csdn.net/txl199106/article/details/62045479

 题意：

　　在一个有向图中，找一个点，使得这个点到其他点的距离和最小，输出距离和，和这个点的坐标。

思路：

　　无根最小树形图，设所有的有向图的距离和为sum。自己建立一个虚拟的原点（n+1），向每一个节点连一条距离为sum+1的边。以n+1为根结点跑一遍最小树形图（复杂度O（VE）），如果求出的ans == -1 或者 ans >=2*（sum + 1），无解，因为这么大的ans，只可能用了两条我们自己建立的边。由于每条边的端点在跑最小树形图的时候会改变，所以记录这是第几条边rtt，结果就是rtt - m，代码中由减了1是因为原图是Base0的。

/*
* @Author: chenkexing
* @Date:   2018-09-05 11:05:14
* @Last Modified by:   chenkexing
* @Last Modified time: 2018-09-10 20:21:22
*/
#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;       
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);



template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/
            const int maxm = 10009;
            const int maxn = 1009;
            struct Edge
            {
                int from,to,c;
            }e[maxm];
            int in[maxn],vis[maxn],pre[maxn],id[maxn];
            int rtt;
            int zhuliu(int root,int n,int m){
                int res = 0;
                while(true){
                    memset(in, inf, sizeof(in));
                    for(int i=1; i<=m; i++){
                        if(e[i].from != e[i].to && e[i].c < in[e[i].to]){
                            pre[e[i].to] = e[i].from;
                            in[e[i].to] = e[i].c;
                            if(e[i].from == root)rtt = i;
                        }
                    }
                    for(int i=1; i<=n; i++){
                        if(i!=root&&in[i] == inf)
                            return -1;
                    }
                    int tn = 0,v;
                    memset(id,-1,sizeof(id));
                    memset(vis,-1,sizeof(vis));

                    in[root] = 0;
                    for(int i=1; i<=n; i++){
                        res += in[i];
                        v = i;
                        while(v !=root && id[v] == -1 && vis[v] != i){
                            vis[v] = i;
                            v = pre[v];
                        }
                        if(v!=root && id[v] == -1){
                            id[v] = ++tn;
                            for(int u = pre[v]; u!=v; u = pre[u]){
                                id[u] = tn;
                            }
                        }
                    }
                    if(tn == 0)break;
                    for(int i=1; i<=n; i++){
                        if(id[i] == -1)id[i] = ++tn;
                    }

                    for(int i=1; i<=m; i++){
                        int v = e[i].to;
                        e[i].to = id[e[i].to];
                        e[i].from = id[e[i].from];
                        if(e[i].to != e[i].from){
                            e[i].c -= in[v];
                        }
                    }

                    n = tn;root = id[root];
                }
                return res;

            }
int main(){
            int n,m,r;
            while(~scanf("%d%d", &n, &m))
            {
                    int sum = 0;
                    for(int i=1; i<=m; i++){
                        int u,v,c;
                        scanf("%d%d%d", &u, &v, &c);
                        u++,v++;
                        e[i].from = u;e[i].to = v;
                        e[i].c = c;
                        sum += c;
                    }
                    sum++;
                    for(int i=m+1; i<=n+m; i++){
                        e[i].from = n+1;
                        e[i].to = i-m;
                        e[i].c = sum;
                    }

                    int ans = zhuliu(n+1,n+1,n+m);
                    // debug(ans);
                    if(ans == -1 || ans - sum >= sum){
                        puts("impossible");
                    }
                    else {
                        printf("%d %d
", ans - sum, rtt - m - 1);
                    }
                    printf("
");
            }


            
            return 0;   
}