UVa4080 Warfare And Logistics

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题目描述传送门

尝试删m条边重新计算n个点的最短路时间复杂度O(nm2logn)
但对于点i只有删除了以i为源点的最短路树上的边才需要重新计算。
预处理好每个点的最短路树，这样最多计算O(n2)次单源最短路，时间复杂度降为O(n2mlogn)
坑点：要用第二小的边代替删去的边。

代码

#include<cstdio>#include<iostream>#include<algorithm>#include<queue>#include<cstring>using namespace std;const int INF=1e9;const int maxn=104,maxm=1005;int h[maxn],d[maxn],sum[maxn],p[maxn],secedge[maxn][maxn],edgeid[maxn][maxn];bool vis[maxn],need[maxn][maxm*2];struct Edge{    int t,d,nxt;    Edge(int a=0,int b=0,int c=0):t(a),d(b),nxt(c){}}edges[maxm*2];int n,m,cnt,L;struct heap{    int d,v;    heap(int a=0,int b=0):d(a),v(b){}    bool operator<(const heap&a)const{        return d>a.d;    }};void addedge(int from,int to,int dist){    edges[++cnt]=Edge(to,dist,h[from]);    h[from]=cnt;    edges[++cnt]=Edge(from,dist,h[to]);    h[to]=cnt;    edgeid[from][to]=edgeid[to][from]=cnt;}int main(){    while(scanf("%d%d%d",&n,&m,&L)!=EOF){        cnt=-1;        memset(h,-1,sizeof(h));        memset(secedge,0,sizeof(secedge));        for(int i=0;i<m;i++){            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            if(!secedge[a][b]){                addedge(a,b,c);                secedge[a][b]=secedge[b][a]=INF;            }            else if(c<edges[edgeid[a][b]].d){                secedge[a][b]=secedge[b][a]=edges[edgeid[a][b]].d;                edges[edgeid[a][b]].d=edges[edgeid[a][b]^1].d=c;            }            else if(c<secedge[a][b]) secedge[a][b]=secedge[b][a]=c;        }        memset(sum,0,sizeof(sum));        memset(need,0,sizeof(need));                int c=0;        for(int i=1;i<=n;i++){            memset(vis,0,sizeof(vis));            priority_queue<heap>q;            q.push(heap(0,i));            for(int i=1;i<=n;i++) d[i]=INF;            d[i]=0;            while(!q.empty()){                heap x=q.top();q.pop();                int u=x.v;                  if(vis[u]) continue;                vis[u]=1;                for(int i=h[u];i>-1;i=edges[i].nxt){                    int v=edges[i].t;                    if(d[u]+edges[i].d<d[v]){                        d[v]=d[u]+edges[i].d;                        p[v]=i;                        q.push(heap(d[v],v));                    }                }            }            for(int j=1;j<=n;j++){                sum[i]+=d[j]==INF?L:d[j];                if(j!=i) need[i][p[j]]=need[i][p[j]^1]=1;            }            c+=sum[i];        }        int ans=0;        for(int i=0;i<2*m;i+=2){            int c1=0;            for(int j=1;j<=n;j++){                if(need[j][i]){                    swap(secedge[edges[i].t][edges[i^1].t],edges[i].d);                    edges[i^1].d=edges[i].d;                    memset(vis,0,sizeof(vis));                    priority_queue<heap>q;                    q.push(heap(0,j));                    for(int i=1;i<=n;i++) d[i]=INF;                    d[j]=0;                    while(!q.empty()){                        heap x=q.top();q.pop();                        int u=x.v;                        if(vis[u]) continue;                        vis[u]=1;                        for(int i=h[u];i>-1;i=edges[i].nxt){                            int v=edges[i].t;                            if(d[u]+edges[i].d<d[v]){                                d[v]=d[u]+edges[i].d;                                q.push(heap(d[v],v));                            }                        }                    }                    for(int k=1;k<=n;k++)                        c1+=d[k]==INF?L:d[k];                    swap(secedge[edges[i].t][edges[i^1].t],edges[i].d);                    edges[i^1].d=edges[i].d;                }                else c1+=sum[j];            }            ans=max(ans,c1);        }        printf("%d %d\n",c,ans);    }    return 0;}

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