AcWing 849. Dijkstra求最短路 I
原题链接
简单
作者:
szywdwd
,
2022-04-21 23:53:18
,
所有人可见
,
阅读 211
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 510;
int n, m;
int g[N][N];// 邻接矩阵
int dist[N];// 1到各点距离
bool st[N];// 集合s
int dijkstra() {
memset(dist, 0x3f, sizeof dist);
dist[1] = 0;
for(int i = 1; i <= n; ++i) {
int t = -1;
for(int j = 1; j <= n; ++j)
if(!st[j] && (t == -1 || dist[t] > dist[j]))
t = j;
st[t] = true;
for(int j = 1; j <= n; ++j)
dist[j] = min(dist[j], dist[t] + g[t][j]);
}
if(dist[n] == 0x3f3f3f3f) return -1;
else return dist[n];
}
int main()
{
cin >> n >> m;
memset(g, 0x3f, sizeof g);// 逐字节填充g,若第二个参数大于1字节,则取低8位来填充g
while(m--) {
int a, b, c;
cin >> a >> b >> c;
g[a][b] = min(g[a][b], c);// 取min是为了去权值较大的重边
}
cout << dijkstra() << endl;
return 0;
}