LA4080/UVa1416 Warfare And Logistics 最短路树

题目大意：

• 求图中两两点对最短距离之和

• 允许你删除一条边，让你最大化删除这个边之后的图中两两点对最短距离之和。

暴力：每次枚举删除哪条边，以每个点为源点做一次最短路，复杂度(O(NM^2logN))。

值得注意的是，(Dijkstra)的复杂度(O(NlogN))是关于边而非点的。

这个复杂度对于(n=100,m=1000)的数据难以接受。我们考虑对每个点建出其最短路树。容易想到，只有删除到这个点的最短路树上的边时，才需要再做一次(Dijkstra)。也就是说每个源点只需要做(n)次最短路，复杂度变成(O(N^2MlogN))。

代码实现起来比较麻烦。。本弱调了整整一晚上。

#include <bits/stdc++.h>
using namespace std;

#define LL long long
const int N = 100 + 5;
const int M = 2000 + 5;
const int INF = 0x3f3f3f3f;

int n, m, l, kase, ban[M];

struct Graph {
	
	int cnt, head[N];
	
	struct Edge {int from, nxt, to, id, w;}e[M];
	
	void clear () {
		cnt = -1;
		for (int i = 1; i <= n; ++i) {
			head[i] = -1;
		}
	}
	
	void add_edge (int u, int v, int w) {
		++cnt; e[cnt] = (Edge) {u, head[u], v, cnt, w}; head[u] = cnt;
	}
	
	struct HeapNode {
		int u; LL d;
		bool operator < (HeapNode rhs) const {
			return d > rhs.d;
		}
	};
	
	priority_queue <HeapNode> pq;
	
	int done[N], _fa[N][N]; LL _dis[N][N]; 
	//dis[i][j] -> i to j
	//fa[i][j] -> i as source, j's father
	
	void dijkstra (int s) {
		kase = kase + 1;
		pq.push ((HeapNode) {s, 0});
		LL *dis = _dis[s]; int *fa = _fa[s];
		for (int i = 1; i <= n; ++i) {
			fa[i] = -1, dis[i] = i == s ? 0 : INF;
		}
		while (!pq.empty ()) {
			HeapNode now = pq.top (); pq.pop ();
			if (done[now.u] == kase) continue;
			for (int i = head[now.u]; ~i; i = e[i].nxt) {
				int v = e[i].to;
				if (ban[i]) continue;//禁用的边 -> 不用 
				if (dis[v] > dis[now.u] + e[i].w) {
					fa[v] = now.u;
					dis[v] = dis[now.u] + e[i].w;
					pq.push ((HeapNode) {v, dis[v]});
				}
			} 
			done[now.u] = kase;
		}
//		cout << "s = " << s << endl;
//		for (int i = 1; i <= n; ++i) {
//			cout << "dis[" << i << "] = " << dis[i] << endl;
//		}
	}
}G;

bool have[N][M]; int minw[N][N];

struct Tree {
	
	vector <int> Gr[N];
	
	int sz[N]; LL sum[N], dis[N];
	
	//sz[u] -> 点u的子树大小
	//sum[u] -> 点u到其子树里所有点的距离和
	
	void prep (int u) {
		sz[u] = 1; sum[u] = dis[u];
		for (int i = 0; i < (int)Gr[u].size (); ++i) {
			int v = Gr[u][i];
			prep (v);  
			sz[u] += sz[v];
			sum[u] += sum[v];
		}
	}
	
	void build (int s, LL *_dis, int *fa, int cmd) {
		for (int i = 1; i <= n; ++i) Gr[i].clear ();
		memcpy (dis, _dis, sizeof (dis));
		for (int i = 1; i <= n; ++i) {
			if (fa[i] != -1) {
				Gr[fa[i]].push_back (i);
				if (cmd == 1) {
					have[fa[i]][i] = true;
					have[i][fa[i]] = true;
				}
			}
		}
		prep (s);
//		for (int i = 1; i <= n; ++i) {
//			cout << "dis[" << i << "] = " << dis[i] << endl;
//			cout << "sum[" << i << "] = " << sum[i] << endl;
//		} 
	} 
	
	LL get_ans (int s, LL *_dis, int *fa, int cmd) {
		build (s, _dis, fa, cmd);
		return sum[s] + (n - sz[s]) * l;
	}
	
}tr[N];//tr[i] -> 点i的最短路树 

signed main () {
//	freopen ("data.in", "r", stdin);
//	freopen ("data.out", "w", stdout);
	while (cin >> n >> m >> l) {
		G.clear ();
		memset (have, 0, sizeof (have));
		memset (minw, 0x3f, sizeof (minw));
		for (int i = 1; i <= m; ++i) {
			static int u, v, w;
			cin >> u >> v >> w;
			G.add_edge (u, v, w);
			G.add_edge (v, u, w);
			minw[u][v] = min (minw[u][v], w);
			minw[v][u] = min (minw[v][u], w);
		}
		LL ans1 = 0, ans2 = 0;
		for (int s = 1; s <= n; ++s) {
			G.dijkstra (s);
			ans1 += tr[s].get_ans (s, G._dis[s], G._fa[s], 1);
			//存一下最初的have 
		}
		cout << ans1 << " ";
		for (int i = 0; i <= G.cnt; i += 2) {
			//每次枚举禁用一条边。
			LL res_now = 0;
			ban[i] = ban[i + 1] = true;//双向都要禁 
			for (int s = 1; s <= n; ++s) { //枚举删除之后每一棵最短路树的状况 
				int u = G.e[i].from, v = G.e[i].to, w = G.e[i].w;
				if (have[u][v] && w == minw[u][v]) G.dijkstra (s);
				res_now += tr[s].get_ans (s, G._dis[s], G._fa[s], 0);
			}
			ban[i] = ban[i + 1] = false;
			ans2 = max (ans2, res_now);
		}
		cout << ans2 << endl; 
	}
}