题目大意
在一个无向图中,定义两个点s,t的最短路径子图为一个极大边集,对于该边集内的所有有向边e,总存在一条起点为s,终点为t且经过边e的路径,使得该路径长度为s到t的最短路径长度。现给出一个无向图,s1, s2, t1, t2四个节点,求一条最长的路径,使得它满足下列条件之一:1. 该路径上的所有边都属于s1到t1的最短路径子图且属于s2到t2的最短路径子图;2. 该路径上的所有边都属于s1到t1的最短路径子图且属于t2到s2的最短路径子图。输出该路径的长度。
题解
我们可以用4次Dijkstra得到s1,t1和s2,t2的最短路径子图$G_1,G_2$,另外由$G_2$可以得到其反向图$G'_2$。然后分别在子图$G_1cap G_2$和$G_1cap G'_2$上进行拓扑排序求最长路径即可。本题最重要的就是从题面到数学语言的翻译过程了,如果这一点不明确,我们可能就会建立一个子图$G_1cap(G_2cup G'_2)$,这样就乱了。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <cassert>
using namespace std;
const int MAX_NODE = 1510, MAX_EDGE = 200005, INF = 0x3f3f3f3f;
struct Node;
struct Edge;
struct Node
{
int DistS, DistT;
bool Done;
int Dist;//Dijkstra时,它是临时的东东;TopSort时,它是真正的最长路径长度
int DfsN;
Edge *Head;
}_nodes[MAX_NODE];
int TotNode;
struct HeapNode
{
Node *cur;
int Dist;
HeapNode(Node *x):cur(x),Dist(x->Dist){}
bool operator < (const HeapNode& a) const
{
return Dist > a.Dist;
}
};
struct Edge
{
int Weight;
bool InSPG;//在第一个人的最短路径子图中
bool InTopG;//在Top子图中
Node *To, *From;
Edge *Next, *Rev;
Edge() :InTopG(false), InSPG(false){}
}_edges[MAX_EDGE];
vector<Edge*> Temp;
int _eCount;
Edge *NewEdge()
{
if (_eCount < MAX_EDGE - 1)
return _edges + ++_eCount;
else
{
Temp.push_back(new Edge());
return Temp.back();
}
}
Edge *AddEdge(Node *from, Node *to, int w)
{
Edge *e = NewEdge();
e->From = from;
e->To = to;
e->Weight = w;
e->Next = from->Head;
from->Head = e;
return e;
}
void Build(int uId, int vId, int w)
{
Node *u = _nodes + uId, *v = _nodes + vId;
Edge *e1 = AddEdge(u, v, w), *e2 = AddEdge(v, u, w);
e1->Rev = e2;
e2->Rev = e1;
}
void Dijkstra(Node *start)
{
static priority_queue<HeapNode> q;
for (int i = 1; i <= TotNode; i++)
{
_nodes[i].Dist = INF;
_nodes[i].Done = false;
}
start->Dist = 0;
q.push(start);
while (!q.empty())
{
HeapNode temp = q.top();
q.pop();
Node *cur = temp.cur;
if (cur->Done)
continue;
cur->Done = true;
for (Edge *e = cur->Head; e; e = e->Next)
{
if (cur->Dist + e->Weight < e->To->Dist)
{
e->To->Dist = cur->Dist + e->Weight;
q.push(e->To);
}
}
}
}
void GetInGraph(void(*DoInGraph)(Edge*), int spLen)
{
for (int i = 1; i <= _eCount; i++)
if (_edges[i].From->DistS + _edges[i].Weight + _edges[i].To->DistT == spLen)
DoInGraph(_edges + i);
for (int i = 0; i < Temp.size(); i++)
if (Temp[i]->From->DistS + Temp[i]->Weight + Temp[i]->To->DistT == spLen)
DoInGraph(Temp[i]);
}
void SetOrgGraph(Edge *e)
{
e->InSPG = true;
}
void MakeTopGraph1(Edge *e)
{
if (e->InSPG)
e->InTopG = true;
}
void MakeTopGraph2(Edge *e)
{
if (e->Rev->InSPG)
e->InTopG = true;
}
int ShortestPath(int s, int t)
{
Node *start = _nodes + s, *target = _nodes + t;
Dijkstra(target);
for (int i = 1; i <= TotNode; i++)
_nodes[i].DistT = _nodes[i].Dist;
Dijkstra(start);
for (int i = 1; i <= TotNode; i++)
_nodes[i].DistS = _nodes[i].Dist;
return target->DistS;
}
void ClearTopGraph()
{
for (int i = 1; i <= _eCount; i++)
_edges[i].InTopG = false;
for (int i = 0; i < Temp.size(); i++)
Temp[i]->InTopG = false;
}
stack<Node*> St;
void Dfs(Node *cur)
{
assert(cur->DfsN != 1);
if (cur->DfsN == 2)
return;
cur->DfsN = 1;
for (Edge *e = cur->Head; e; e = e->Next)
{
if (!e->InTopG)
continue;
Dfs(e->To);
}
cur->DfsN = 2;
St.push(cur);
}
int LongestPath()
{
for (int i = 1; i <= TotNode; i++)
{
_nodes[i].Dist = 0;
_nodes[i].DfsN = 0;
}
for (int i = 1; i <= TotNode; i++)
Dfs(_nodes + i);
int ans = 0;
while (!St.empty())
{
Node *cur = St.top();
St.pop();
ans = max(ans, cur->Dist);
for (Edge *e = cur->Head; e; e = e->Next)
if (e->InTopG)
e->To->Dist = max(e->To->Dist, cur->Dist + e->Weight);
}
return ans;
}
int main()
{
int totEdge, s1, s2, t1, t2;
scanf("%d%d%d%d%d%d", &TotNode, &totEdge, &s1, &t1, &s2, &t2);
for (int i = 1; i <= totEdge; i++)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
Build(u, v, w);
}
int len1 = ShortestPath(s1, t1);
GetInGraph(SetOrgGraph, len1);
int len2 = ShortestPath(s2, t2);
GetInGraph(MakeTopGraph1, len2);
int ans1 = LongestPath();
ClearTopGraph();
GetInGraph(MakeTopGraph2, len2);
int ans2 = LongestPath();
printf("%d
", max(ans1, ans2));
return 0;
}