百度之星 初赛三 最短路 2 Dijkstra

打比赛的时候切的，不过竟然 wa 了 14 次~
挺简单的，直接在跑 $Dijkstra$ 的时候记录一下路径最大值就好了.

#include <bits/stdc++.h>
#define inf 100000000000000
#define ll long long
#define mod 998244353
#define N 1003
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
int n,m,s;
struct Node
{
    int u;
    ll dis;
    Node(int u=0,ll dis=0):u(u),dis(dis){}
    bool operator<(Node b)const
    {
        return b.dis<dis;
    }
};
priority_queue<Node>q;
int hd[N],nex[N<<2],pre[N<<2],edges,done[N],to[N<<2],now[N<<2];
ll f[N][N],d[N];
ll val[N<<2];
inline void addedge(int u,int v,ll c)
{
    nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;
}
inline void Dijkstra()
{
    int i,u,v;
    for(i=1;i<=n;++i) done[i]=0;
    for(i=0;i<=n;++i) d[i]=inf,pre[i]=0,now[i]=i;
    d[s]=0, q.push(Node(s,0)), pre[s]=0,now[s]=0;
    while(!q.empty())
    {
        Node e=q.top(); u=e.u,q.pop();
        if(done[u]) continue;
        done[u]=1;
        if(u!=s)
        {
            now[u]=max(u, pre[u]);
        }
        for(i=hd[u];i;i=nex[i]){
            if(d[to[i]]>=d[u]+val[i]){
                if(d[to[i]]>d[u]+val[i])
                {
                    d[to[i]]=d[u]+val[i];
                    pre[to[i]]=now[u];
                    q.push(Node(to[i],d[to[i]]));
                }
                else {
                    if(now[u]<pre[to[i]]) pre[to[i]]=now[u];
                  }
            }
        }
    }
}
int main() {
    setIO("input");
    using namespace IO;
    int T;
    scanf("%d",&T);
    while(T--) {
        int i,j;
        scanf("%d%d",&n,&m);
        edges=0;
        for(i=1;i<=n;++i) hd[i]=0;
        for(i=1;i<=n;++i) for(j=1;j<=n;++j) f[i][j]=inf;
        for(i=1;i<=n;++i) f[i][i]=0;
        for(i=1;i<=m;++i) {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            addedge(a,b,(ll)c), addedge(b,a,(ll)c), f[a][b]=f[b][a]=min(f[a][b],(ll)c);
        }
        int ans=0;
        for(i=1;i<=n;++i) {
            s=i, Dijkstra();
            for(j=1;j<=n;++j)
                 ans=(long long) (ans+pre[j])%mod;
        }
        printf("%d\n",ans);   

    }
    return 0;
}