hdu 6714 最短路 2 floyd dijkstra

至少得知道floyd的最初形式,dp[k][i][j]，只经过前k个点，i到j的最短路。

所以题目就是问你，任意两点间的最短路，经过的最大编号的点最小是多少，还不含两个端点。

魔改一下dijkstra就行了，每个点跑一次。先比较长度，长度相同比较下最大编号谁的更小。每个点存的经过最大编号并不包含这个点本身。记得特判下起点终点直连的情况。

#include <cstdio>
#include <queue> 
using namespace std;
typedef long long ll;
const int mo = 998244353,MAXN = 1010,MAXM = 4010;
const ll inf = 1e14;
struct pot
{
    int x,maxn;
    ll dis;
    pot (int _x = 0,ll _dis = 0,int _maxn = 0) : x(_x),dis(_dis),maxn(_maxn) {}
    friend bool operator < (pot a,pot b)
    {
        if (a.dis != b.dis)
            return a.dis > b.dis;
        return a.maxn > b.maxn;
    }
};
struct dat
{
    int maxn;
    ll dis;
    dat (ll _dis = 0,int _maxn = 0) : dis(_dis),maxn(_maxn) {}
    friend bool operator < (dat a,dat b)
    {
        if (a.dis != b.dis)
            return a.dis > b.dis;
        return a.maxn > b.maxn;
    }
};
priority_queue <pot> que;
dat dis[MAXN];
int head[MAXN],to[MAXM],nxt[MAXM],val[MAXM];
int T,n,m,cnt,ans;
bool vis[MAXN];
void add(int x,int y,int v)
{
    nxt[++cnt] = head[x];
    to[cnt] = y;
    head[x] = cnt;
    val[cnt] = v;
}
void dijkstra(int s)
{
    for (int i = 1;i <= n;i++)
    {
        dis[i] = dat(inf,0);
        vis[i] = false;
    }
    que.push(pot(s,0,0));
    dis[s] = dat(0,0);
    while(que.empty() == false)
    {
        pot now = que.top();
        que.pop();
        if(vis[now.x] == true) 
            continue;
        vis[now.x] = true;
        for(int i = head[now.x]; i; i = nxt[i])
        {
            if (now.x != s)
            {
                if(dis[to[i]].dis > dis[now.x].dis + val[i] || (dis[to[i]].dis == dis[now.x].dis + val[i]) && (max(dis[now.x].maxn,now.x) < dis[to[i]].maxn))
                {
                    dis[to[i]].dis = dis[now.x].dis + val[i];
                    dis[to[i]].maxn = max(dis[now.x].maxn,now.x);
                    que.push(pot(to[i],dis[to[i]].dis,dis[to[i]].maxn));
                }
            }else
            {
                if(dis[to[i]].dis >= dis[now.x].dis + val[i])
                {
                    dis[to[i]].dis = dis[now.x].dis + val[i];
                    dis[to[i]].maxn = 0;
                    que.push(pot(to[i],dis[to[i]].dis,dis[to[i]].maxn));
                }
            }
        }
    }
}
int main()
{
    for (scanf("%d",&T);T != 0;T--)
    {
        for (int i = 1;i <= cnt;i++)
            nxt[i] = 0;
        for (int i = 1;i <= n;i++)
            head[i] = 0;
        cnt = ans = 0;
        scanf("%d%d",&n,&m);
        int tx,ty,tv;
        for (int i = 1;i <= m;i++)
        {
            scanf("%d%d%d",&tx,&ty,&tv);
            add(tx,ty,tv);
            add(ty,tx,tv);
        }
        for (int i = 1;i <= n;i++)
        {
            dijkstra(i);
            for (int j = 1;j <= n;j++)
                ans = ((ll)ans + dis[j].maxn) % mo;
        }
        printf("%d
",ans);
    }
    return 0;
}