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 << "\n"; #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 '\n' #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\n", ans - sum, rtt - m - 1); } printf("\n"); } return 0; }