题意:
N<=1000,M<=10000,不定根最小树形图,输出最小权和相应的根位置
分析:
设一个虚根,然后连虚边,权值为一个大于总权和的权但是小于INF,用边标号来记录实根的位置
如果存在最小树形图的话,只会有一条虚边,换句话说实根是唯一的
代码:
//
// Created by TaoSama on 2015-10-31
// Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 2e4 + 10;
int n, m;
struct Edge {
int u, v, c;
} e[M];
int realRoot, pre[N], in[N], id[N], vis[N];
int directedMST(int root, int n, int m) {
int ret = 0;
while(true) {
memset(in, 0x3f, sizeof in);
//find minimum "in" edge
for(int i = 0; i < m; ++i) {
int u = e[i].u, v = e[i].v;
if(u != v && e[i].c < in[v]) {
if(u == root) realRoot = i;
pre[v] = u;
in[v] = e[i].c;
}
}
for(int i = 0; i < n; ++i) {
if(i == root) continue;
if(in[i] == INF) return -1; //isolate vertex
}
int cnt = 0;
memset(id, -1, sizeof id);
memset(vis, -1, sizeof vis);
in[root] = 0;
for(int i = 0; i < n; ++i) {
ret += in[i];
int u = i; //judge circle
while(id[u] == -1 && vis[u] != i && u != root) {
vis[u] = i;
u = pre[u];
}
//relabel
if(id[u] == -1 && u != root) {
for(int v = pre[u]; v != u; v = pre[v])
id[v] = cnt;
id[u] = cnt++;
}
}
if(cnt == 0) break; //no circle, already MST
for(int i = 0; i < n; ++i) //other vertex
if(id[i] == -1) id[i] = cnt++;
//reconstruct graph
for(int i = 0; i < m; ++i) {
int u = e[i].u, v = e[i].v;
e[i].u = id[u];
e[i].v = id[v];
if(u != v) e[i].c -= in[v];
}
n = cnt;
root = id[root];
}
return ret;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d%d", &n, &m) == 2) {
int sum = 0;
for(int i = 0; i < m; ++i) {
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].c);
sum += e[i].c;
}
++sum;
for(int i = 0; i < n; ++i)
e[m + i] = (Edge) {n, i, sum};
int ans = directedMST(n, n + 1, n + m);
if(ans == -1 || ans - sum >= sum) puts("impossible");
else printf("%d %d\n", ans - sum, realRoot - m);
puts("");
}
return 0;
}