题目大意

给你一个图，每条边有一个代价，让你求0到1在最短路径的前提下的最小代价

---

解题思路

bfs同时求个最代价

---

代码

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 200
using namespace std;
int n, m, x, y, z, tot, b[N], p[N], v[N], head[N];
queue<int>d;
struct rec
{
	int to, l, next;
}a[N*10];
void add(int x, int y, int z)
{
	a[++tot].to = y;
	a[tot].l = z;
	a[tot].next = head[x];
	head[x] = tot;
	return;
}
void bfs()
{
	memset(b, 127/3, sizeof(b));
	memset(v, 127/3, sizeof(v));
	b[0] = 0;
	v[0] = 0;
	p[0] = 1;
	d.push(0);
	while(!d.empty())//bfs
	{
		int h = d.front();
		d.pop();
		for (int i = head[h]; i; i = a[i].next)
			if (b[h] + 1 < b[a[i].to] || b[h] + 1 == b[a[i].to] && v[h] + a[i].l < v[a[i].to])//没到过或长度一样且代价更小
			{
				b[a[i].to] = b[h] + 1;
				v[a[i].to] = v[h] + a[i].l;
				if (!p[a[i].to])
				{
					d.push(a[i].to);
					p[a[i].to] = 1;
				}
			}
	}
	return;
}
int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; ++i)
	{
		scanf("%d%d%d", &x, &y, &z);
		add(x, y, z);
	}
	bfs();
	printf("%d", v[1]);
	return 0;
}