SRM 584 div1 lev3（最小樹形圖）

題目鏈接

閒來無聊在TC上搜羅圖論題目， 無意中發現了這題， 此題構圖其實挺好想，我一開始想的是費用流

後來構完圖才發現是個最小樹形圖。。。 只是覺得div1 lev3的題目不應只有這種程度吧。。。。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

class FoxTheLinguist {
public:
	int minimalHours(int, vector <string>);
};

typedef int LL;
const LL INF = (int)1e9;
const int N = 105;
const int M = 1005;

struct DMST {
	struct Edge {
		int u, v;
		LL w;

		void init(int a, int b, LL c) {
			this->u = a, this->v = b, w = c;
		}
	};

	Edge E[M];
	LL in[N];
	int no[N], pre[N], vis[N];
	int n, m, rt;

	void init(int n, int r) {
		this->n = n;
		rt = r;
		m = 0;
	}

	void add(int u, int v, LL w) {
		E[m].init(u, v, w);
		m++;
	}

	LL gao() {
		LL res = 0;
		while (1) {
			fill(in, in + n, INF);
			for (int i = 0; i < m; i++) {
				int u = E[i].u, v = E[i].v;
				if (u != v && E[i].w < in[v]) {
					in[v] = E[i].w;
					pre[v] = u;		
				}
			}

			for (int i = 0; i < n; i++) {
				if (i == rt) continue;
				if (in[i] == INF) return -1;
			}
			
			int cnt = 0;
			fill(no, no + n, -1);
			fill(vis, vis + n, -1);
			
			in[rt] = 0;
			for (int i = 0; i < n; i++) {
				res += in[i];
				int v = i;
				while (vis[v] != i && no[v] == -1 && v != rt) {
					vis[v] = i;
					v = pre[v];
				}

				if (v != rt && no[v] == -1) {
					for (int u = pre[v]; u != v; u = pre[u]) {
						no[u] = cnt;
					}
					no[v] = cnt++;
				}	
			}

			if (cnt == 0) break;
			for (int i = 0; i < n; i++)
				if (no[i] == -1) {
					no[i] = cnt++;
				}

			for (int i = 0; i < m; i++) {
				int v = E[i].v;
				E[i].u = no[E[i].u];
				E[i].v = no[E[i].v];
				if (E[i].u != E[i].v)
					E[i].w -= in[v];
			}

			n = cnt;
			rt = no[rt];
					
		}

		return res;		
	}	
}G;

int FoxTheLinguist::minimalHours(int a, vector<string> info) {
	int n = a * 10 + 1;
	int s = n - 1;
	G.init(n, s);

	int m = info.size();
	string tmp = "";

	for (int i = 0; i < m; i++)
		tmp += info[i];

	int id = 0;
	int sz = tmp.size();
	

	while (id < sz) {
		if (tmp[id] == ' ') id++;
		
		int u = (tmp[id] - 'A') * 10 + tmp[1 + id] - '0';
		int v = (tmp[4 + id] - 'A') * 10 + tmp[5 + id] - '0';

		int w = 0;

		for (int j = 7; j <= 10; j++)
			w += (tmp[j + id] - '0'), w *= 10;
		w /= 10;
			
		G.add(u, v, w);

		id += 11;
	}

	for (int i = 0; i < a; i++) 
		G.add(s, i * 10, 0);

	for (int i = 0; i < a; i++) {
		int st = (i + 1) * 10 - 1;
		for (int j = 0; j < 9; j++) {
			G.add(st, st - 1, 0);
			st--;
		}
	}

	return G.gao();
}

<%:testing-code%>
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