奇怪的连通图_1545

题目描述：

已知一个无向带权图，求最小整数k。使仅使用权值小于等于k的边，节点1可以与节点n连通。

输入：

输入包含多组测试用例，每组测试用例的开头为一个整数n(1 <= n <= 10000)，m(1 <= m <= 100000)，代表该带权图的顶点个数，和边的个数。
接下去m行，描述图上边的信息，包括三个整数,a(1 <= a <= n),b(1 <= b <= n),c(1 <= c <= 1000000),表示连接顶点a和顶点b的无向边，其权值为c。

输出：

输出为一个整数k，若找不到一个整数满足条件，则输出-1。

样例输入：

3 31 3 51 2 32 3 23 21 2 32 3 53 11 2 3 

样例输出：

35-1

二分查找 复杂度较高，其实考察的是并查集

#include <iostream>#include <vector>#include <algorithm>#include <cstdio>using namespace std; struct edge_t {    int a, b;    int dist;}; int n, m;vector<edge_t> edges;vector<int> father; int findFather(int x) {    while (x != father[x]) {        x = father[x];    }    return x;} bool cmp(const edge_t &a, const edge_t &b) {    return a.dist < b.dist;} void init() {    for (int i = 0; i < father.size(); ++i)         father[i] = i;    sort(edges.begin(), edges.end(), cmp);} void solve() {    for (int i = 0; i < m; ++i) {        int fa = findFather(edges[i].a);        int fb = findFather(edges[i].b);        if (fa < fb) father[fb] = fa;        else father[fa] = fb;        if (findFather(n) == 1) {            cout << edges[i].dist << endl;            return;        }    }    cout << "-1" << endl;} int main() {    while (scanf("%d %d", &n, &m) != EOF) {        father.resize(n + 1);        edges.resize(m);        for (int i = 0; i < m; ++i) {            scanf("%d %d %d", &edges[i].a, &edges[i].b, &edges[i].dist);        }        init();        solve();    }    return 0;}
也可以用c语言实现

#include <stdio.h>#include <stdlib.h>#include <stdlib.h> #define N 20005 typedef struct node {    int bt, ed, value;} node; int father[N]; int findSet(int x){    int root;     if (x == father[x]) {        return x;    }     root = findSet(father[x]);    father[x] = root;     return root;} int cmp(const void *p, const void *q){    const node *a = p;    const node *b = q;     return a->value - b->value;}  int main(void){    int i, n, m, pa, pb, flag;    node *nodes;     while (scanf("%d %d", &n, &m) != EOF) {        // 初始化并查集        for (i = 1; i <= n; i ++) {            father[i] = i;        }         nodes = (node *)malloc(sizeof(node) * m);         for (i = 0; i < m; i ++) {            scanf("%d %d %d", &nodes[i].bt, &nodes[i].ed, &nodes[i].value);        }        qsort(nodes, m, sizeof(nodes[0]), cmp);         for (i = flag = 0; i < m; i ++) {            pa = findSet(nodes[i].bt);            pb = findSet(nodes[i].ed);             if (pa != pb) {                father[pa] = pb;            }             if (findSet(1) == findSet(n)) {                flag = 1;                printf("%d\n", nodes[i].value);                break;            }        }         if (! flag)            printf("-1\n");         free(nodes);    }         return 0;}

另外也可以用BFS

然后将队列改为优先队列果断A了

#include<cstdio>#include<cstring>#include<vector>#include<queue>using namespace std;struct node{    int ne;    int va;};struct node1{int next;int m;bool operator < (const node1& a) const{return m>a.m;}};vector<node> map[10002];int visit[10002];int n,t,in;int mmin(int aa,int bb){    return aa<bb?aa:bb;}int max(int aa,int bb){    return aa>bb?aa:bb;}int BFS(){int xiao = 1000008;priority_queue<node1> qu;struct node1 needru,needpush,var;needru.next = 1;needru.m = 0;qu.push(needru);visit[1] = 1;while(!qu.empty()){needpush = qu.top();qu.pop();if(needpush.next==n)xiao = mmin(xiao,needpush.m);int len = map[needpush.next].size();visit[needpush.next] = 1;for(int i=0;i<len;i++){if(!visit[map[needpush.next][i].ne]){var.next = map[needpush.next][i].ne;var.m = max(map[needpush.next][i].va,needpush.m);qu.push(var);}}}return xiao;}int main(){    int i,j,a,b;    struct node qq;    while(scanf("%d%d",&n,&t)!=EOF)    {        in = 1000008;        memset(visit,0,sizeof(visit));        for(i=0;i<t;i++)        {            scanf("%d%d%d",&a,&b,&j);            qq.ne = b;            qq.va = j;            map[a].push_back(qq);            qq.ne = a;            map[b].push_back(qq);        }        visit[1] = 1;in = BFS();        if(in!=1000008)        printf("%d\n",in);        else            printf("-1\n");        for(i=0;i<=n;i++)        map[i].clear();    }     return 0;}