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

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解题思路

bfs同时求个最代价

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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;
}