【朴素Dijkstra】AcWing849.Dijkstra求最短路 I

AcWing849.Dijkstra求最短路 I

题解

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N = 510, M = 1e5 + 10;


int d[N], n, m;
bool flag[N];
int g[N][N]; //朴素只能用邻接矩阵不然会变成O(nm)


void Dijkstra()
{
    memset(d, 0x3f, sizeof d);
    d[1] = 0;
    for(int i = 0; i < n; ++i)
    {
        int mi = 0x3f3f3f, k;
        for(int j = 1; j <= n; ++j)
            if(!flag[j] && d[j] < mi)
                mi = d[j], k = j;
        flag[k] = true;
        for(int j = 1; j <= n; ++j)
            if(!flag[j] && mi + g[k][j] < d[j])
                d[j] = mi + g[k][j];
    }
}

int main()
{
    memset(g, 0x3f, sizeof g);
    int x, y, z;
    scanf("%d%d",&n,&m);
    for(int i = 0 ;i < m; ++i)
    {
        scanf("%d%d%d",&x,&y,&z);
        g[x][y] = min(g[x][y], z);
    }
    Dijkstra();
    if(flag[n])
    cout << d[n] << endl;
    else
        cout << -1 << endl;
    return 0;
}

标签：int,短路,Dijkstra,AcWing849,include,朴素
来源： https://www.cnblogs.com/czy-algorithm/p/16307350.html