题目大意
给出一个有边权和点权的无向图,规定路径长度为路径经过的边权和。求出这么一条从起点到终点的路径,它的长度小于等于给出的长度限制(条件1),且经过的点权最大的点的点权是所有满足条件1的路径中的最小的。
题解
图论题我没怎么接触过需要用到反演思想的,然而这道题就是。我们发现这个图有这么一个单调性:选的点数越多,子图中起点到终点的最短路径就会越短。因此我们每次枚举一个节点,将点权小于等于该节点的节点都纳入子图(显然该节点的点权越大,加入子图中的节点数越多),看看图中的最短路径长度是否小于等于长度限制。显然枚举的节点的点权越大,满足限制条件可能性越大,反之越小。这样我们就可以二分了。
#include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; const int MAX_NODE = 10010, MAX_EDGE = 100010, INF = 0x3f3f3f3f; int DistLimit; struct Node; struct Edge; struct Node { Edge *Head; int Weight, Dist; bool Done, InGraph; }_nodes[MAX_NODE], *Sorted[MAX_NODE]; int TotNode; struct Edge { Node *To; Edge *Next; int Weight; }_edges[MAX_EDGE]; int _eCount; struct CmpPNode { bool operator () (Node *a, Node *b) { return a->Dist > b->Dist; } }; bool Cmp(Node *a, Node *b) { return a->Weight < b->Weight; } void AddEdge(Node *from, Node *to, int w) { Edge *e = _edges + ++_eCount; e->To = to; e->Weight = w; e->Next = from->Head; from->Head = e; } void Build(int uId, int vId, int w) { Node *u = _nodes + uId, *v = _nodes + vId; AddEdge(u, v, w); AddEdge(v, u, w); } struct HeapNode { long long Dist; Node *CurNode; HeapNode(long long dist, Node *curNode):Dist(dist), CurNode(curNode){} bool operator < (const HeapNode& a) const { return Dist > a.Dist; } }; void Dijkstra() { for (int i = 1; i <= TotNode; i++) { _nodes[i].Dist = INF; _nodes[i].Done = false; } if (!_nodes[1].InGraph) return; _nodes[1].Dist = 0; static priority_queue<HeapNode> q; q.push(HeapNode(0, _nodes + 1)); while (!q.empty()) { HeapNode temp = q.top(); q.pop(); Node *cur = temp.CurNode; if (cur->Done) continue; cur->Done = true; for (Edge *e = cur->Head; e; e = e->Next) { if (!e->To->InGraph) continue; if (cur->Dist + e->Weight < e->To->Dist) { e->To->Dist = cur->Dist + e->Weight; q.push(HeapNode(e->To->Dist, e->To)); } } } } bool Alive(int x) { for (int i = 1; i <= x; i++) Sorted[i]->InGraph = true; for (int i = x + 1; i <= TotNode; i++) Sorted[i]->InGraph = false; Dijkstra(); return _nodes[TotNode].Dist <= DistLimit; } int LowerBound(int l, int r, bool (*InRange) (int)) { if (!InRange(r)) return -1; while (l < r) { int mid = (l + r) / 2; if (InRange(mid)) r = mid; else l = mid + 1; } return l; } int main() { int totEdge; scanf("%d%d%d", &TotNode, &totEdge, &DistLimit); for (int i = 1; i <= TotNode; i++) scanf("%d", &_nodes[i].Weight); for (int i = 1; i <= totEdge; i++) { int u, v, w; scanf("%d%d%d", &u, &v, &w); Build(u, v, w); } for (int i = 1; i <= TotNode; i++) Sorted[i] = _nodes + i; sort(Sorted + 1, Sorted + TotNode + 1, Cmp); int p = LowerBound(1, TotNode, Alive); if (p == -1) printf("AFK "); else printf("%d ", Sorted[p]->Weight); return 0; }