HUD.2544 最短路 (Dijkstra)

HUD.2544 最短路 (Dijkstra)

题意分析

1. 1表示起点，n表示起点(或者颠倒过来也可以)
2. 建立无向图
3. 从n或者1跑dij即可。

代码总览

#include <bits/stdc++.h>
#define nmax 110
#define inf 1e8
using namespace std;
int mp[nmax][nmax];
int shortpath[nmax];
bool visit[nmax];
int n,m;

void dij(int s)
{
    int cur = s;
    memset(visit,0,sizeof(visit));
    for(int i = 1;i<=n;++i) shortpath[i] = inf;
    shortpath[cur] = 0;
    visit[cur] = true;
    for(int i = 1;i<=n ;++i){
        for(int j = 1;j<=n;++j){
            if(!visit[j] && shortpath[cur] + mp[cur][j] < shortpath[j]){
                shortpath[j] = shortpath[cur] + mp[cur][j];
            }
        }
        int nowmin = inf;
        for(int j = 1; j<=n;++j){
            if(!visit[j] && shortpath[j] <nowmin){
                nowmin = shortpath[j];
                cur = j;
            }
        }
        visit[cur] = true;
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d %d",&n,&m)!=EOF){
        if( n == 0 && m == 0) break;
        int sta,ed,cost;
        for(int i = 1;i<=n;++i){
            for(int j = 1;j<=n;++j){
                mp[i][j] = inf;
            }
        }
        for(int i = 0; i<m;++i){
            scanf("%d %d %d",&sta,&ed,&cost);
            int temp = mp[sta][ed];
            if(cost < temp){
                mp[sta][ed] = mp[ed][sta] = cost;
            }
        }
        dij(1);
        printf("%d
",shortpath[n]);
    }
    return 0;
}