Codeforces Alpha Round #20 (Codeforces format) C. Dijkstra?(裸的dijkstra)

题目链接：http://codeforces.com/problemset/problem/20/C

思路：需要用优化过的dijkstra，提供两种写法。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#define REP(i, a, b) for (int i = (a); i < (b); ++i)
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define TR(iter, c) for (__typeof(c.begin()) iter = c.begin(); iter != c.end(); ++iter)
using namespace std;

const int MAX_N = (100000 + 100);
const long long inf = 1LL << 60;
int N, M, pre[MAX_N];
long long dist[MAX_N];

struct cmp {
	bool operator() (const int &p1, const int &p2) const {
		return dist[p1] <= dist[p2];
	}
};
vector<pair<int, int > > g[MAX_N];
set<int, cmp> q;

void Dijkstra()
{
	dist[1] = 0;
	q.insert(1);
	while (!q.empty()) {
		int u = *q.begin();
		q.erase(q.begin());
		REP(i, 0, (int)g[u].size()) {
			int v = g[u][i].first, w = g[u][i].second;
			if (dist[u] + w < dist[v]) {
				q.erase(v);
				dist[v] = dist[u] + w;
				pre[v] = u;
				q.insert(v);
			}
		}
	}
}

void dfs(int u)
{
	if (u != 1) dfs(pre[u]);
	printf("%d ", u);
}
int main()
{
	cin >> N >> M;
	FOR(i, 1, N) dist[i] = inf;
	FOR(i, 1, M) {
		int u, v, w; cin >> u >> v >> w;
		g[u].push_back(make_pair(v, w));
		g[v].push_back(make_pair(u, w));
	}
	Dijkstra();
	if (dist[N] >= inf) { puts("-1"); return 0; }
	dfs(N);
	return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#define REP(i, a, b) for (int i = (a); i < (b); ++i)
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
using namespace std;

const int MAX_N = (100000 + 100);
const long long inf = 1LL << 60;
int N, M, pre[MAX_N], vis[MAX_N];
long long dist[MAX_N];

struct cmp {
	bool operator() (const pair<long long, int>&p, const pair<long long, int>&q) const {
		return p.first >= q.first;
	}
};
vector<pair<int, int > > g[MAX_N];
priority_queue<pair<long long, int>, vector<pair<long long, int> >, cmp> pq;

void Dijkstra()
{
	pq.push(make_pair(dist[1] = 0, 1));
	while (!pq.empty()) {
		pair<long long, int> p = pq.top();
		pq.pop();
		if (vis[p.second]) continue;
		vis[p.second] = 1;
		REP(i, 0, (int)g[p.second].size()) {
			int v = g[p.second][i].first, w = g[p.second][i].second;
			if (p.first + w < dist[v]) {
				dist[v] = p.first + w;
				pre[v] = p.second;
				if (!vis[v]) pq.push(make_pair(dist[v], v));
			}
		}

	}
}

void dfs(int u)
{
	if (u != 1) dfs(pre[u]);
	printf("%d ", u);
}
int main()
{
	cin >> N >> M;
	FOR(i, 1, N) dist[i] = inf, vis[i] = 0;
	FOR(i, 1, M) {
		int u, v, w; cin >> u >> v >> w;
		g[u].push_back(make_pair(v, w));
		g[v].push_back(make_pair(u, w));
	}
	Dijkstra();
	if (dist[N] >= inf) { puts("-1"); return 0; }
	dfs(N);
	return 0;
}