uvalive 4080 Warfare And Logistics最短路树

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求删去一条边后，任意两点距离之和最大值。

对于每个节点它到其他点的最短路径存在一个最短路树，若不破坏这个最短路树，那么它到其余点的最短路径不变。所以本题只需对每个节点求n次最短路。

#include<iostream>#include<string>#include<cstring>#include<cstdio>#include<cmath>#include<iomanip>#include<map>#include<algorithm>#include<queue>#include<set>#define inf 1000000000000#define pi acos(-1.0)#define eps 1e-8#define seed 131using namespace std;typedef pair<int,int> pii;typedef unsigned long long ull;typedef long long ll;const int maxn=100005;int n,m;struct Edge{    int from,to,w;    Edge(){}    Edge(int a,int b,int c):from(a),to(b),w(c){}};struct node{    int w,id;    node(int a,int b):w(a),id(b){}    bool operator<(const node& a)const    {        return w>a.w;    }};ll l;bool tree[105][2005];vector<Edge>edge;vector<int>g[105];ll cost[105];ll d[105];void dij(int s);ll dij2(int s,int p);int main(){    int a,b,c;    while(~scanf("%d%d%lld",&n,&m,&l))    {        edge.clear();        for(int i=1;i<=n;i++)            g[i].clear();        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&a,&b,&c);            edge.push_back(Edge(a,b,c));            edge.push_back(Edge(b,a,c));            g[a].push_back(edge.size()-2);            g[b].push_back(edge.size()-1);        }        memset(tree,0,sizeof(tree));        for(int i=1;i<=n;i++)            dij(i);        ll ans=0;        for(int i=1;i<=n;i++)            ans+=cost[i];        cout<<ans;        ll mx=0;        ll y;        for(int i=0;i<m*2;i+=2)        {            y=0;            for(int j=1;j<=n;j++)            {                if(tree[j][i]==1||tree[j][i+1]==1)                {                    y+=dij2(j,i);                }                else                    y+=cost[j];            }            mx=max(mx,y);        }        cout<<" "<<mx<<endl;    }    return 0;}void dij(int s){    bool vis[105];    int p[105];    memset(p,-1,sizeof(p));    memset(vis,0,sizeof(vis));    for(int i=1;i<=n;i++)        d[i]=inf;    d[s]=0;    priority_queue<node>que;    que.push(node(0,s));    while(!que.empty())    {        node u=que.top();        que.pop();        int len=g[u.id].size();        for(int i=0;i<len;i++)        {            Edge& e=edge[g[u.id][i]];            if(vis[e.to]==0&&d[e.to]>u.w+e.w)            {                d[e.to]=u.w+e.w;                p[e.to]=g[u.id][i];                que.push(node(d[e.to],e.to));            }        }    }    for(int i=1;i<=n;i++)    {        if(i==s||p[i]==-1)            continue;        tree[s][p[i]]=1;    }    cost[s]=0;    for(int i=1;i<=n;i++)    {        if(d[i]==inf)            cost[s]+=l;        else            cost[s]+=d[i];    }}ll dij2(int s,int p){    bool vis[105];    memset(vis,0,sizeof(vis));    for(int i=1;i<=n;i++)        d[i]=inf;    d[s]=0;    priority_queue<node>que;    que.push(node(0,s));    while(!que.empty())    {        node u=que.top();        que.pop();        vis[u.id]=1;        int len=g[u.id].size();        for(int i=0;i<len;i++)        {            if(g[u.id][i]==p||g[u.id][i]==p+1)                continue;            Edge& e=edge[g[u.id][i]];            if(vis[e.to]==0&&d[e.to]>u.w+e.w)            {                d[e.to]=u.w+e.w;                que.push(node(d[e.to],e.to));            }        }    }    ll ans=0;    for(int i=1;i<=n;i++)    {        if(d[i]==inf)            ans+=l;        else            ans+=d[i];    }    return ans;}

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