poj_3662 最小化第k大的值

题目大意

有N个节点以及连接的P个无向边，现在要通过这P条边从1号节点连接到N号节点。若无法连接成功，则返回-1；若能够连接成功，那么其中用到了L条边，这L条边中有K条边可以免费，L-K条边不能免费，求出不能免费的边的最大长度。

题目分析

判断能否到达，可以通过BFS搜索路径，若不能到达，返回-1；若能到达，且最少需要的路径的边数小于等于K，那么所有的边都可以免费，则返回0；若能够到达，且最少需要的路径边数大于K，则需要求出从节点1到节点N的路径中第K+1长的边的最小值，即最小化第k大的值问题。 
    用二分法，枚举边长x，若路径中大于等于x的边数大于K，则说明x在构成路径的边长序列中的序号大于K+1，则将x增大；若路径中大于等于x的边长小于等于K，则说明x在构成路径的边长序列中的序号小于等于K+1，因此将x减小.... 直到x为最小的满足路径中边长大于等于x的边数大于K的最小值，则x即为路径中第K+1大的边，且X最小。 
    那么，对于一个长度x，该选择哪条路径呢？由于要求的最小的x，那么就要求路径中边长大于等于x的个数尽可能的少（若个数少于等于K，则减小x），转换一下，将路径中长度大于等于x的边权值设为1，小于x的权值设为0，则走一条从源到汇的路径，路径中边的权值和最小即对应路径中边长大于等于x的个数最少。 
    求最短路径，使用Dijkstra算法即可。

实现(c++)

#include<stdio.h>
#include<string.h>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
#define MAX_NODE 1005
#define MAX_EDGE 20005
#define INF 1 << 28
#define min(a, b) a < b?a:b
#define max(a, b) a >b? a:b
struct Edge{
	int to;
	int d;
	int next;
};
struct NodeDist{
	int v;
	int d;
	NodeDist(int vv = 0, int dd = 0) :
		v(vv), d(dd){};
};
struct Cmp{
	bool operator()(const NodeDist& e1, const NodeDist& e2){
		return e1.d > e2.d;
	}
};
Edge gEdges[MAX_EDGE];
int gEdgeCount;
int gHead[MAX_NODE];
int gDist[MAX_NODE];
bool gVisited[MAX_NODE];
void InsertEdge(int u, int v, int d){
	int e = gEdgeCount++;
	gEdges[e].to = v;
	gEdges[e].d = d;
	gEdges[e].next = gHead[u];
	gHead[u] = e;

	e = gEdgeCount++;
	gEdges[e].to = u;
	gEdges[e].d = d;
	gEdges[e].next = gHead[v];
	gHead[v] = e;
}

int Dijkstra(int s, int t, int k){
	memset(gVisited, false, sizeof(gVisited));
	memset(gDist, 0x7F, sizeof(gDist));
	gDist[s] = 0;
	priority_queue<NodeDist, vector<NodeDist>, Cmp> pq;
	pq.push(NodeDist(s, 0));
	NodeDist nd;
	while (!pq.empty()){
		nd = pq.top();
		pq.pop();
		if (gVisited[nd.v])
			continue;
		gVisited[nd.v] = true;
		if (nd.v == t){
			break;
		}
		for (int e = gHead[nd.v]; e != -1; e = gEdges[e].next){
			int v = gEdges[e].to;
			if (gDist[v] > gDist[nd.v] + (gEdges[e].d >= k)){
				gDist[v] = gDist[nd.v] + (gEdges[e].d >= k);
				pq.push(NodeDist(v, gDist[v]));
			}
		}
	}
	return gDist[t];
}

int MinStep(int s, int t){
	bool visited[MAX_NODE];
	memset(visited, false, sizeof(visited));
	queue<pair<int, int> > Q;
	Q.push(pair<int, int>(s, 0));
	visited[s] = true;
	while (!Q.empty()){
		pair<int, int> p = Q.front();
		Q.pop();
		if (p.first == t){
			return p.second;
		}
		for (int e = gHead[p.first]; e != -1; e = gEdges[e].next){
			if (!visited[gEdges[e].to]){
				Q.push(pair<int, int>(gEdges[e].to, p.second + 1));
				visited[gEdges[e].to] = true;
			}
		}
	}
	return -1;
}

int gEdgeDist[MAX_EDGE];
int main(){
	int n, p, k, u, v, d, e_count;
	while (scanf("%d %d %d", &n, &p, &k) != EOF){
		memset(gHead, -1, sizeof(gHead));
		gEdgeCount = 0;
		e_count = 0;
		for (int i = 0; i < p; i++){
			scanf("%d %d %d", &u, &v, &d);
			InsertEdge(u, v, d);
			gEdgeDist[e_count++] = d;
		}

		int min_step = MinStep(1, n);
		if (min_step == -1){
			printf("-1\n");
			continue;
		}
		else if (min_step <= k){
			printf("0\n");
			continue;
		}

		sort(gEdgeDist, gEdgeDist + e_count);
		int beg = 0, end = e_count, mid;
		int rr;
		while (beg < end){
			mid = (beg + end) / 2;
			int t = Dijkstra(1, n, gEdgeDist[mid]);
			if (t <= k){
				end = mid;
			}
			else{
				rr = gEdgeDist[mid];
				beg = mid + 1;
			}
		}
		//最后得到的是，满足路径中边长大于等于 x 的长度的边数 大于k 最大的 x

		printf("%d\n", rr);
	}
	return 0;
}