前言
题目很水 就是跑一遍 最短路
传送门 :
思路使用vector存边 然后开始跑
CODE#include
#include
#include
#include
#define x first
#define y second
using namespace std;
typedef pair pii;
int n,m,s,t;
const int N = 3000,M = 6500;
int dist[N];
int st[N];
struct edge{
int to,val;
};
vector g[N];
int dij()
{
memset(dist,0x3f,sizeof dist);
dist[s] = 0;
priority_queue heap;
heap.push({0,s});
while(heap.size())
{
pii k = heap.top();
heap.pop();
int ver = k.y ,distance = k.x;
if(st[ver]) continue;
st[ver] = true;
for(auto x : g[ver])
{
int j = x.to;
if(dist[j] > distance + x.val)
{
dist[j] = distance + x.val;
heap.push({dist[j],j});
}
}
}
return dist[t];
}
void solve()
{
cin>>n>>m>>s>>t;
while(m -- )
{
int a,b,c;
cin>>a>>b>>c;
g[a].push_back({b,c});
g[b].push_back({a,c});
}
int ans = dij();
cout
关注
打赏