POJ3463 Sightseeing

题目大意：求两点间最短路与长度为最短路长度+1的路径的条数之和。

方法1：最短路径+DP

首先求出ST间最短路径，然后根据递归式记忆化搜索（因此还要构造反向图）。

我们知道到达终点的路径长度最长为maxDist(T)=minDist(T)+1，而与终点相连的节点的路径长度最长为maxDist(T)-Weight(e)。这些节点与更前面的节点也是如此。于是我们从终点开始递归，利用PathCnt(v)=sum(PathCnt(u)) (PathCnt(u)=|{Dist(u)|Dist(u)<=maxDist(u)}|)。

初值注意不能直接PathCnt(S)设为1，因为S点也可能在一个环内。因此我们应当建点S'->S，边权为0。

代码：

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

#define LOOP(i,n) for(int i=1; i<=n; i++)
const int MAX_NODE = 1010, MAX_EDGE = 10010 * 2, INF = 0x3f3f3f3f;

struct Node;
struct Edge;

struct Node
{
	int Id, Dist, PathCnt[2];
	bool Inq;
	Edge *Head, *RevHead;
}_nodes[MAX_NODE], *Start, *Target;
int _vCount;

struct Edge
{
	int Weight;
	Node *From, *To;
	Edge *Next;
}*_edges[MAX_EDGE];
int _eCount;

void Init(int vCount)
{
	_vCount = vCount;
	_eCount = 0;
	memset(_nodes, 0, sizeof(_nodes));
}

void SetST(int sId, int tId)
{
	Start = sId + _nodes;
	Target = tId + _nodes;
}

Edge *NewEdge()
{
	_eCount++;
	return _edges[_eCount] ? _edges[_eCount] : _edges[_eCount] = new Edge();
}

void AddEdge(Node *from, Node *to, int weight, bool IsRev)
{
	Edge *e = NewEdge();
	e->From = from;
	e->To = to;
	e->Weight = weight;
	Edge *&head = IsRev ? from->RevHead : from->Head;
	e->Next = head;
	head = e;
}

void Build(int uId, int vId, int weight)
{
	Node *from = uId + _nodes, *to = vId + _nodes;
	from->Id = uId;
	to->Id = vId;
	AddEdge(from, to, weight, false);
	AddEdge(to, from, weight, true);
}

void SPFA()
{
	static queue<Node*> q;
	LOOP(i, _vCount)
	{
		_nodes[i].Dist = INF;
		_nodes[i].Inq = false;
	}
	Start->Dist = 0;
	Start->Inq = true;
	q.push(Start);
	while (!q.empty())
	{
		Node *u = q.front();
		q.pop();
		u->Inq = false;
		for (Edge *e = u->Head; e; e = e->Next)
		{
			if (u->Dist + e->Weight < e->To->Dist)
			{
				e->To->Dist = u->Dist + e->Weight;
				if (!e->To->Inq)
				{
					e->To->Inq = true;
					q.push(e->To);
				}
			}
		}
	}
}

int CntPath(Node *cur, int maxDist)
{
	if (cur->Dist > maxDist)
		return 0;
	int &pathCnt = cur->PathCnt[maxDist - cur->Dist];
	if (pathCnt != -1)
		return pathCnt;
	if (cur == Start)
		return 1;
	pathCnt = 0;
	for (Edge *e = cur->RevHead; e; e = e->Next)
		pathCnt += CntPath(e->To, maxDist - e->Weight);
	return pathCnt;
}

int main()
{
#ifdef _DEBUG
	freopen("c:\\noi\\source\\input.txt", "r", stdin);
#endif
	int testCase, totNode, totEdge, uId, vId, weight, sId, tId;
	scanf("%d", &testCase);
	while (testCase--)
	{
		scanf("%d%d", &totNode, &totEdge);
		Init(totNode + 1);
		LOOP(i, totEdge)
		{
			scanf("%d%d%d", &uId, &vId, &weight);
			Build(uId, vId, weight);
		}
		scanf("%d%d", &sId, &tId);
		Build(totNode + 1, sId, 0);
		SetST(totNode + 1, tId);
		SPFA();
		LOOP(i, _vCount)
			_nodes[i].PathCnt[0] = _nodes[i].PathCnt[1] = -1;
		printf("%d\n", CntPath(Target, Target->Dist + 1));
	}
	return 0;
}

方法2：Dijkstra

同时更新一个节点的最短路径长度和次短路长度以及它们的个数。优先队列里维护节点，key值为节点最短路长度或次短路长度。究竟是哪个长度我们不用管它，该节点能更新最短路就更新最短路，否则看看能不能更新次短路。

注意：1.如果v的newDist与v原来的路径长度相等，更新完路径个数后不要入队！队列只维护最短路径或次短路径。2.更新完了v最短路，v次短路也更新了，所以要把v节点以key值为最短路长度和次短路长度入队两次。3.优先队列是大根堆。

代码：

#include <cstdio>
#include <cstring>
#include <cassert>
#include <queue>
using namespace std;

#define LOOP(i,n) for(int i=1; i<=n; i++)
#define _1st 0
#define _2nd 1
const int MAX_NODE = 1010, MAX_EDGE = 10010 * 2, INF = 0x3f3f3f3f;

struct Node;
struct Edge;

struct Node
{
	int Id, Dist[2], Cnt[2];
	bool Done[2];
	Edge *Head;
}_nodes[MAX_NODE], *Start, *Target;
int _vCount;

struct Edge
{
	int Weight;
	Node *From, *To;
	Edge *Next;
}*_edges[MAX_EDGE];
int _eCount;

struct HeapNode
{
	Node *Org;
	int Dist;
	bool Is2nd;
	bool operator <(const HeapNode a)const { return Dist > a.Dist; }
	HeapNode(Node *a, bool isSec):Org(a),Dist(a->Dist[isSec]),Is2nd(isSec){}
};

void Init(int vCount)
{
	memset(_nodes, 0, sizeof(_nodes));
	_vCount = vCount;
	_eCount = 0;
}

void SetST(int sId, int tId)
{
	Start = sId + _nodes;
	Target = tId + _nodes;
}

Edge *NewEdge()
{
	_eCount++;
	if (_edges[_eCount])
		return _edges[_eCount];
	else
		return _edges[_eCount] = new Edge();
}

Edge *AddEdge(Node *from, Node *to, int weight)
{
	Edge *e = NewEdge();
	e->From = from;
	e->To = to;
	e->Weight = weight;
	e->Next = e->From->Head;
	e->From->Head = e;
	return e;
}

void Build(int uId, int vId, int weight)
{
	Node *u = uId + _nodes, *v = vId + _nodes;
	u->Id = uId;
	v->Id = vId;
	AddEdge(u, v, weight);
}

void Dijkstra()
{
	static priority_queue<HeapNode> q;
	LOOP(i, _vCount)
	{
		_nodes[i].Dist[0] = _nodes[i].Dist[1] = INF;
		_nodes[i].Done[0] = _nodes[i].Done[1] = false;
	}
	Start->Dist[_1st] = 0;
	Start->Cnt[_1st] = 1;
	q.push(HeapNode(Start, false));
	while (!q.empty())
	{
		HeapNode cur = q.top();
		q.pop();
		Node *u = cur.Org;
		if (u->Done[cur.Is2nd])
			continue;
		u->Done[cur.Is2nd] = true;
		for (Edge *e = u->Head; e; e = e->Next)
		{
			int newDist = cur.Dist + e->Weight;
			if (newDist < e->To->Dist[_1st])
			{
				e->To->Cnt[_2nd] = e->To->Cnt[_1st];
				e->To->Dist[_2nd] = e->To->Dist[_1st];
				e->To->Cnt[_1st] = u->Cnt[cur.Is2nd];
				e->To->Dist[_1st] = newDist;
				q.push(HeapNode(e->To, _1st));
				q.push(HeapNode(e->To, _2nd));
			}
			else if (newDist > e->To->Dist[_1st] && newDist < e->To->Dist[_2nd])
			{
				e->To->Cnt[_2nd] = u->Cnt[cur.Is2nd];
				e->To->Dist[_2nd] = newDist;
				q.push(HeapNode(e->To, _2nd));
			}
			else if (newDist == e->To->Dist[_1st])
				e->To->Cnt[_1st] += u->Cnt[cur.Is2nd];
			else if (newDist == e->To->Dist[_2nd])
				e->To->Cnt[_2nd] += u->Cnt[cur.Is2nd];
		}
	}
}

int main()
{
#ifdef _DEBUG
	freopen("c:\\noi\\source\\input.txt", "r", stdin);
	freopen("c:\\noi\\source\\output.txt", "w", stdout);
#endif
	int testCase, totNode, totEdge, uId, vId, weight, sId, tId;
	scanf("%d", &testCase);
	while (testCase--)
	{
		scanf("%d%d", &totNode, &totEdge);
		Init(totNode);
		LOOP(i, totEdge)
		{
			scanf("%d%d%d", &uId, &vId, &weight);
			Build(uId, vId, weight);
		}
		scanf("%d%d", &sId, &tId);
		SetST(sId, tId);
		Dijkstra();
		printf("%d\n", Target->Cnt[_1st] + ((Target->Dist[_2nd] == Target->Dist[_1st] + 1) ? Target->Cnt[_2nd] : 0));
	}
	return 0;
}

　　错误做法：SPFA同时更新最短路和次短路。

为什么Dijkstra就可以？因为u出队时，由于Dijkstra的贪心，u本身就更新完了，无后顾之忧；而SPFA中，u还没更新完就要往下更新。u往下更新以后，如果以后运算过程中u自己再被更新，此时再由u往下更新v，v节点就错了。