题目链接
题意
n个点,m条边,求从1到n的路径上最大的最小边,(如,有1 2 3,1 3 4, 2 3 5这几条边,从1到3最大的最小边就是4)
思路
松弛操作替换为d[v] = min(d[u], dis[u][v]),因为要找最大的边,所以每次进行松弛的的时候是找最大的边。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <queue>
#define pr pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1e5 + 7;
typedef long long ll;
int n, m, head[maxn], to[maxn<<1], edge[maxn<<1], nex[maxn<<1], cnt;
int d[maxn], vis[maxn];
void add(int x, int y, int val) {
to[cnt] = y;
edge[cnt] = val;
nex[cnt] = head[x];
head[x] = cnt++;
}
void init() {
memset(head, -1, sizeof(head));
cnt = 0;
}
void dijkstra(int s) {
memset(vis, 0, sizeof(vis));
memset(d, 0, sizeof(d));//*
priority_queue<pr, vector<pr>, less<pr> > que;//*
que.push(make_pair(0, s));
d[s] = INF;//*
while (!que.empty()) {
int u = que.top().second;
que.pop();
vis[u] = 1;
for (int i = head[u]; ~i; i = nex[i]) {
int v = to[i], len = edge[i];
if(!vis[v] && d[v] < min(d[u], len))//*
{
d[v] = min(d[u], len);//*
que.push(make_pair(d[v], v));
}
}
}
}
int main()
{
int tt, Case = 1;
cin >> tt;
while (tt--) {
cin >> n >> m;
init();
for (int i = 1; i <= m; i++) {
int x, y, z;
scanf("%d %d %d", &x, &y, &z);
add(x, y, z); add(y, x, z);
}
dijkstra(1);
printf("Scenario #%d:\n", Case++);
printf("%d\n\n", d[n]);
}
}