UVALive4080[Warfare And Logistics] 最短路树+dijkstra

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题目描述 | OJ链接

题意：①先求任意两点间的最短路径累加和，其中不连通的边权为L ②删除任意一条边,求全局最短路径和的最大值。

解题报告：由于要求[多源最短路][路径和]，我们首先可以想到[Floyed]，但是有删边的操作，所以[Floyed]时间复杂度为O(n*n*n*m), 直接跑dijkstra复杂度为o(n*m*m*logn), 复杂度也很大，所以我们可以使用[最短路树]，只有要删的边在树上时，我们才需要重新跑[dijkstra]，由于树上有n-1条边，所以复杂度为o(n*n*m*logn);

ps:要开long long

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 105;const int M = 1000;const long long Inf = 1e12 + 7;#define LL long long LL l;struct Edge{    int u, v, w, id;    Edge(int u,int v,int w,int id):u(u),v(v),w(w),id(id){}};struct HeapNode{    LL dist;    int u;    bool operator<(const HeapNode& rhs) const{        return dist>rhs.dist;    }};struct Dijkstra{    int n, m;    vector<Edge> edges;    vector<int> G[N];    LL d[N], ans, ret, dt[N];    bool done[N];    int p[N], link[M][N], del[M];    void init(int n){        memset(dt,0,sizeof(dt));        memset(link,0,sizeof(link));         ans=ret=0;        this->n=n;        for ( int i=1; i<=n; i++ ) G[i].clear();        edges.clear();      }    void addeage(int u, int v, int w, int id){        edges.push_back(Edge(u,v,w,id));        m=edges.size();        G[u].push_back(m-1);    }    void dijkstra1(int s){        for ( int i=1; i<=n; i++ ) d[i]=Inf, done[i]=p[i]=0;        priority_queue<HeapNode> Q;        Q.push((HeapNode){0,s});        d[s]=0;        while( !Q.empty() ){            HeapNode x=Q.top(); Q.pop();            int u=x.u;            if( done[u] ) continue;            done[u]=1;            link[p[u]][s]=1;            for ( int i=0; i<G[u].size(); i++ ){                Edge& e=edges[G[u][i]];                if( d[e.v]>d[u]+e.w ){                    d[e.v]=d[u]+e.w;                    p[e.v]=e.id;                    Q.push((HeapNode){d[e.v],e.v});                }            }        }        for ( int i=1; i<=n; i++ ){            if( d[i]==Inf ) ans+=l, dt[s]+=l;            else ans+=d[i], dt[s]+=d[i];        }    }    LL dijkstra2(int s){        for ( int i=1; i<=n; i++ ) d[i]=Inf, done[i]=0;        d[s]=0;        priority_queue<HeapNode> Q;        Q.push((HeapNode){0,s});        while( !Q.empty() ){            HeapNode x=Q.top(); Q.pop();            int u=x.u;            if( done[u] ) continue;            done[u]=1;            for ( int i=0; i<G[u].size(); i++ ){                Edge& e=edges[G[u][i]];                if( del[e.id] ) continue;                if( d[e.v]>d[u]+e.w ){                    d[e.v]=d[u]+e.w;                    Q.push((HeapNode){d[e.v],e.v});                }            }        }        LL tt=0;        for ( int i=1; i<=n; i++ ){            if( d[i]==Inf ) tt+=l;            else tt+=d[i];        }           return tt;     }};Dijkstra D;int main(){    int n, m;    while( scanf("%d%d%lld", &n, &m, &l )==3 ){        D.init(n);        for ( int i=1; i<=m; i++ ){            int u, v, w;            scanf("%d%d%d", &u, &v, &w );            D.addeage(u, v, w, i);            D.addeage(v, u, w, i);        }        for ( int i=1; i<=n; i++ ) D.dijkstra1(i);    //  for ( int i=1; i<=n; i++ ) printf("%lld\n", D.dt[i]);    /*  for ( int i=1; i<=m; i++ ){            for ( int j=1; j<=n; j++ ) printf("%d ", D.link[i][j] );            printf("\n");         }    */        for ( int i=1; i<=m; i++ ){            D.del[i]=1;            LL tt=0;              for ( int j=1; j<=n; j++ )                 if( D.link[i][j] ) tt+=D.dijkstra2(j);                else tt+=D.dt[j];    //      printf("%lld\n", tt);            D.del[i]=0;            D.ret=max(D.ret,tt);        }        printf("%lld %lld\n", D.ans, D.ret );    }}

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