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  • 在加权无向图上求出一条从1号结点到N号结点的路径,使路径上第K+1大的边权尽量小

    二分+最短路算法

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    #include<vector>
    #define maxn 100010
    using namespace std;
    const int INF = 0x3f3f3f3f;
    struct Node {
    	int p;
    	int len;
    	Node(int a, int b) :p(a), len(b) {}
    };
    vector<Node>G[maxn];
    void insert(int be, int en, int len) {
    	G[be].push_back(Node(en, len));
    }
    bool operator <(const Node a, const Node b) {
    	return a.len > b.len;
    }
    int vis[maxn];
    int dis[maxn];
    int n, m, k;
    int dijstra(int be, int range) {
    	memset(vis, 0, sizeof(vis));
    	memset(dis, INF, sizeof(dis));
    	priority_queue<Node>que;
    	que.push(Node(be, 0));
    	dis[be] = 0;
    	while (!que.empty()) {
    		Node ans = que.top();
    		que.pop();
    		if (vis[ans.p]) continue;
    		vis[ans.p] = 1;
    		int x = ans.p;
    		for (int i = 0; i < G[x].size(); i++) {
    			int p = G[x][i].p;
    			int len;
    			if (G[x][i].len >= range) len = 1;
    			else len = 0;
    
    			if (dis[p] > dis[x] + len) {
    				dis[p] = dis[x] + len;
    				que.push(Node(p, dis[p]));
    			}
    		}
    	}
    	return dis[n];
    }
    int check(int mid) {
    	int len = dijstra(1, mid);
    	if (len >= k + 1) return 0;
    	else return 1;
    }
    int main() {
    	int be, en, len;
    	scanf("%d%d%d", &n, &m, &k);
    	for (int i = 0; i < m; i++) {
    		scanf("%d%d%d", &be, &en, &len);
    		insert(be, en, len);
    		insert(en, be, len);
    	}
    	int l = 0;
    	int r = 10000000;
    	int mid;
    	int flag = 0;
        
    	while (r - l > 1) {
    		mid = (r + l) / 2;
    		if (check(mid)) {//往小了压
    			r = mid;
    		}
    		else {
    			l = mid ;
    		}
    	}
    	if (r == 10000000) cout << "-1" << endl;
    	else cout << l << endl;
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/lesning/p/11695427.html
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