hiho_1139_二分+bfs搜索

题目

    给定N个点和M条边，从点1出发，到达点T。寻找路径上边的个数小于等于K的路径，求出所有满足条件的路径中最长边长度的最小值。 
题目链接:二分 
    最小化最大值，考虑采用二分搜索。对所有的边长进行排序，二分，对每次选择的边长，判断是否存在一条路径满足路径上所有的边长度均小于等于该边长，且路径中的边的个数小于等于K。 
    在判断路径是否存在的时候，使用BFS搜索，因为BFS用于寻找最短（边的个数最少）路径。

实现

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#include<stdio.h> #include<string.h> #include<iostream> #include<string> #include<set> #include<map> #include<vector> #include<queue> #include<stack> #include<unordered_map> #include<unordered_set> #include<algorithm> using namespace std; struct Edge{     int to;     int dist;     int next; }; //边的个数，开始的时候数组开的长度为 100005， hihocoder上提示 TLE，实际应该为 RE！！！ //数组开到 200005，就AC了 Edge gEdges[200005]; int gHead[10005]; bool gVisited[10005]; int gEdgeIndex; void Init(){     gEdgeIndex = 0;     memset(gEdges, -1, sizeof(gEdges));     memset(gHead, -1, sizeof(gHead));     memset(gVisited, false, sizeof(gVisited)); } void InsertEdge(int u, int v, int d){     int e = gEdgeIndex++;     gEdges[e].to = v;     gEdges[e].dist = d;     gEdges[e].next = gHead[u];     gHead[u] = e; } struct Node{     int id;     int step;     Node(int i = 0, int s = 0) :id(i), step(s){}; }; //BFS 搜索，寻找从点1到达点boss的路径，要求路径上的所有边的长度都小于等于max_d，且路径长度最大为k //判断能否找到满足要求的路径 bool Arrive2Boss(int boss, int max_d, int k){     queue<Node> Q;     Node node(1, 1);     Q.push(node);     memset(gVisited, false, sizeof(gVisited));     while (!Q.empty()){         node = Q.front();         Q.pop();         if (node.id == boss)             return true;         if (gVisited[node.id])             continue;         gVisited[node.id] = true;         for (int e = gHead[node.id]; e != -1; e = gEdges[e].next){             int v = gEdges[e].to;             if (!gVisited[v] && gEdges[e].dist <= max_d && node.step <= k){                 Q.push(Node(v, node.step + 1));             }         }     }     return false; } int edges_len[100005]; int main(){     int N, M, K, T, u, v, d;     Init();     scanf("%d %d %d %d", &N, &M, &K, &T);     for (int i = 0; i < M; i++){         scanf("%d %d %d", &u, &v, &d);         InsertEdge(u, v, d);         InsertEdge(v, u, d);         edges_len[i] = d;     }     //对路径进行排序     sort(edges_len, edges_len + M);     int beg = 0, end = M;     //二分查找     while (beg < end){         int mid = (beg + end) / 2;         int max_d = edges_len[mid];         if (Arrive2Boss(T, max_d, K)){             end = mid;         }         else             beg = mid + 1;     }     int result = edges_len[beg];     printf("%d\n", result);     return 0; }