SDUT 2498-AOE网上的关键路径(spfa+字典序路径)

9 11
1 2 6
1 3 4
1 4 5
2 5 1
3 5 1
4 6 2
5 7 9
5 8 7
6 8 4
8 9 4
7 9 2

18
1 2
2 5
5 7
7 9

最长路+记录字典序最小路径（即假设有多条最长路输出字典序最小的那条 比方 1->2->4 和 1->3->4 都符合最长路，那么输出1->2->4 ） 主要实现就是在松弛时，当dis[v]==dis[u]+w 时，推断一下路径的字典序来决定是否更新路径，眼下还是仅仅会暴力推断QAQ

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <vector>
#include <string>
#include <cmath>
#include <map>
#include <queue>
using namespace std;
#define LL long long
const int INF = 0x3f3f3f3f;
int s1[10010], s2[10010], ans[10010], dis[10010], in[10010], out[10010], path[10010], n, m, s, e;
bool vis[10010];
vector <pair<int, int> > eg[50010];
bool ok(int u, int v)
{
	int p = v, num1 = 0;
	s1[num1++] = v;

	while (path[p] != -1) {
		s1[num1++] = path[p];
		p = path[p];
	}

	p = u;
	int num2 = 0;
	s2[num2++] = v;
	s2[num2++] = u;

	while (path[p] != -1) {
		s2[num2++] = path[p];
		p = path[p];
	}

	int i = num1 - 1, j = num2 - 1;

	while (i >= 0 && j >= 0) {
		if (s1[i] > s2[j]) {
			return 1;
		}

		i--;
		j--;
	}

	return 0;
}
void spfa()
{
	queue <int> Q;

	for (int i = 1; i <= n; i++) {
		dis[i] = -INF;
	}

	dis[s] = 0;
	Q.push(s);

	while (!Q.empty()) {
		int u = Q.front();
		Q.pop();
		vis[u] = 0;

		for (int i = 0; i < eg[u].size(); i++) {
			int v = eg[u][i].first;
			int w = eg[u][i].second;

			if (dis[v] < dis[u] + w) {
				dis[v] = dis[u] + w;
				path[v] = u;

				if (!vis[v]) {
					vis[v] = 1;
					Q.push(v);
				}
			} else
				if (dis[v] == dis[u] + w && ok(u, v)) {

					path[v] = u ;

					if (!vis[v]) {
						vis[v] = 1;
						Q.push(v);
					}
				}
		}
	}

}
void print()
{
	int p = e, num = 0;

	while (path[p] != -1) {
		ans[num++] = path[p];
		p = path[p];
	}

	printf("%d\n", dis[e]);

	for (int i = num - 1; i > 0; i--) {
		printf("%d %d\n", ans[i], ans[i - 1]);
	}

	printf("%d %d\n", ans[0], e);
}
int main()
{
	int u, v, c;

	while (~scanf("%d %d", &n, &m)) {
		for (int i = 0; i <= n; i++) {
			eg[i].clear();
		}

		memset(in, 0, sizeof(in));
		memset(out, 0, sizeof(out));
		memset(vis, 0, sizeof(vis));
		memset(path, -1, sizeof(path));

		while (m--) {
			scanf("%d%d%d", &u, &v, &c);
			eg[u].push_back(make_pair(v, c));
			in[v]++;
			out[u]++;
		}

		for (int i = 1; i <= n; i++) {
			if (!in[i]) {
				s = i;
			}

			if (!out[i]) {
				e = i;
			}
		}

		spfa();
		print();
	}

	return 0;
}