【Dijkstra模板】codeforces715B Complete The Graph(最短路径)

记录一下Dijkstra的模板，主要加了一个优先队列，时间复杂度变成O(mlogn)，速度快了很多。

struct Edge
{
	int from, to, dist;
	Edge(int u, int v, int w) :from(u), to(v), dist(w) {}
};
struct HeapNode
{
	int d, u;
	HeapNode(int x, int y) :d(x), u(y) {}
	bool operator<(const HeapNode &rhs) const
	{
		return d > rhs.d;
	}
};
struct Dijkstra
{
	vector<Edge> edges;		//边保存
	vector<int> G[maxn];	//保存每个点能到达的所有边
	bool done[maxn];		//是否已经永久标记
	int d[maxn];			//起点到各个点的距离
	int p[maxn];			//最短路的上一条弧
	void init()
	{
		for (int i = 0; i < maxn; i++)
			G[i].clear();
		edges.clear();
	}
	void addedge(int from, int to, int dist)
	{
		edges.push_back(Edge(from, to, dist));
		G[from].push_back(edges.size());
	}
	void dij()
	{
		priority_queue<HeapNode> Q;
		for (int i = 0; i < n; i++) d[i] = INF;
		d[s] = 0;
		memset(done, 0, sizeof(done));
		Q.push(HeapNode(0,s));
		while (!Q.empty())
		{
			HeapNode x = Q.top(); Q.pop();
			int u = x.u;
			if (done[u]) continue;
			done[u] = true;
			for (int i = 0; i < G[u].size(); i++)
			{
				Edge& e = edges[G[u][i]];
				if (d[e.to] > d[u] + e.dist)
				{
					d[e.to] = d[u] + e.dist;
					p[e.to] = G[u][i];
					Q.push(HeapNode(d[e.to], e.to));
				}
			}
		}
	}
};

下面写一下cf上这道题：

题意：

给出一个无向图，现在要求将所有边权为0的边的权值变成大于等于1的值，问怎么变化使最后的最短路径和为l

要点：

基本就是一个最短路径，题目想起来不难，先把所有权值不为0的图进行一次dij，特判一下。然后将权值为0的边一条一条赋值为1放入图并进行dij，如果此时的最短路径d<=l，那就把这条边改成l - dij()+w即可，剩余所有边改成无穷大。但这题我WA了快10次，主要是输出很烦，而且一开始模板还写错，数组上限也开小了，还是太菜了。

#include<iostream>
#include<cstring>
#include<string>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1005;
const int maxm = 20010;//注意这里
const ll INF = 1000000000000000000;
int n, m, l, s, t, cnt,tot;
int S[maxm], T[maxm];

struct Edge
{
	int from, to, dist;
	Edge(int u, int v, int w) :from(u), to(v), dist(w) {}
};
struct HeapNode
{
	int d, u;
	HeapNode(int x, int y) :d(x), u(y) {}
	bool operator<(const HeapNode &rhs) const
	{
		return d > rhs.d;
	}
};
struct Dijkstra
{
	vector<Edge> edges;		//边保存
	vector<int> G[maxn];	//保存每个点能到达的所有边
	bool done[maxn];		//是否已经永久标记
	ll d[maxn];			//起点到各个点的距离
	int p[maxn];			//最短路的上一条弧
	void init()
	{
		for (int i = 0; i < maxn; i++)
			G[i].clear();
		edges.clear();
	}
	void addedge(int from, int to, int dist)
	{
		edges.push_back(Edge(from, to, dist));
		G[from].push_back(edges.size() - 1);
	}
	ll dij()
	{
		priority_queue<HeapNode> Q;
		for (int i = 0; i < maxn; i++) d[i] = INF;
		d[s] = 0;
		memset(done, 0, sizeof(done));
		Q.push(HeapNode(0, s));
		while (!Q.empty())
		{
			HeapNode x = Q.top(); Q.pop();
			int u = x.u;
			if (done[u]) continue;
			done[u] = true;
			for (int i = 0; i < G[u].size(); i++)
			{
				Edge& e = edges[G[u][i]];
				if (d[e.to] > d[u] + e.dist&&d[u]+e.dist<=l)//这里有个优化
				{
					d[e.to] = d[u] + e.dist;
					p[e.to] = G[u][i];
					Q.push(HeapNode(d[e.to], e.to));
				}
			}
		}
		return d[t];
	}
	void print()
	{
		printf("YES\n");
		int t = edges.size();
		for (int i = 0; i < t - 2; i += 2)
			printf("%d %d %d\n", edges[i].from, edges[i].to, edges[i].dist);
		//printf("------------\n");
		printf("%d %d %d\n", edges[t - 2].from, edges[t - 2].to, l - dij() + edges[t - 2].dist);
		//printf("------------\n");
		for (int i = tot; i<=cnt; i++)
			printf("%d %d %I64d\n", S[i], T[i], INF);
	}
};
Dijkstra graph;

int main()
{
	//while (1)
	{
		scanf("%d%d%d%d%d", &n, &m, &l, &s, &t);
		int u, v, w;
		graph.init();
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d%d", &u, &v, &w);
			if (w)
			{
				graph.addedge(u, v, w);
				graph.addedge(v, u, w);
			}
			else
			{
				S[++cnt] = u;
				T[cnt] = v;
			}
		}
		tot = 1;
		if (graph.dij() < l)//一开始特判
		{
			//printf("%d\n", graph.dij());
			printf("NO\n");
			return 0;
		}
		else if (graph.dij() == l)
		{
			graph.print();
			return 0;
		}
		for (; tot <= cnt; tot++)//将每个边分别赋值为1加入，如果最短路径小于l就脱出
		{
			graph.addedge(S[tot], T[tot], 1);
			graph.addedge(T[tot], S[tot], 1);
			if (graph.dij() <= l)
				break;
		}
		tot++;
		//printf("%d %d\n", tot, cnt);
		if (tot > cnt + 1)
			printf("NO\n");
		else
			graph.print();
	}
	return 0;
}