给一个无向图,求出两个值,所有点到所有其他点的最短距离和,任意删除一条边后的这个值。
数据规模是100点1000边。
白书例题,不多说了直接对于每个点求出最短路树,对于每条边,如果它不是最短路树上的边,那么我们不需要对它进行最短路计算了,由于点数只有100,那么树上的边只有n-1,所以我们对于以每个点为源点的树,只需要重新计算n-1次最短路即可,每次计算复杂度为n*m,最终复杂度就是n*n*m*log()。个人觉得有点高,不过实际跑起来还是很快的。
注意有重边,要多加判断了。
召唤代码君:
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #define maxn 2222 typedef long long ll; using namespace std; struct heapnode{ ll D,U; bool operator < (heapnode ND) const { return D>ND.D; } }; ll inf=~0U>>2; ll to[maxn],c[maxn],next[maxn],first[maxn],edge; ll u[maxn],v[maxn],w[maxn],minlen[maxn][maxn],tim[maxn][maxn]; ll dis[maxn],from[maxn],C[maxn],f[maxn][maxn]; bool key[maxn],akey[maxn],done[maxn]; ll n,m,L,ans,sum; void _init() { edge=-1,sum=ans=0; for (int i=1; i<=n; i++) { first[i]=-1,C[i]=0; for (int j=1; j<=n; j++) minlen[i][j]=inf; for (int j=1; j<=m; j++) f[i][j]=0; } } void addedge(int U,int V,int W) { edge++; to[edge]=V,c[edge]=W,next[edge]=first[U],first[U]=edge; edge++; to[edge]=U,c[edge]=W,next[edge]=first[V],first[V]=edge; } ll dijkstra(int S,int EG,ll Dis[],ll From[],bool Key[]) { priority_queue<heapnode> Q; for (int i=1; i<=n || i<=m; i++) { if (i<=n) Dis[i]=inf,From[i]=-1,done[i]=false; if (i<=m) Key[i]=false; } Dis[S]=0; Q.push((heapnode){0,S}); while (!Q.empty()) { heapnode ND=Q.top(); Q.pop(); int cur=ND.U; if (done[cur]) continue; else done[cur]=true; if (From[cur]!=-1) Key[From[cur]/2+1]=true; for (int i=first[cur]; i!=-1; i=next[i]) if (i!=EG+EG-2 && i!=EG+EG-1 && Dis[cur]+c[i]<Dis[to[i]]) { Dis[to[i]]=Dis[cur]+c[i]; From[to[i]]=i; Q.push((heapnode){Dis[to[i]],to[i]}); } } ll tot=0; for (int i=1; i<=n; i++) if (Dis[i]!=inf) tot+=Dis[i]; else tot+=L; return tot; } int main() { inf*=inf; while (scanf("%lld%lld%lld",&n,&m,&L)!=EOF) { _init(); for (int i=1; i<=m; i++) { scanf("%lld%lld%lld",&u[i],&v[i],&w[i]); addedge(u[i],v[i],w[i]); if (w[i]<minlen[u[i]][v[i]]) minlen[u[i]][v[i]]=minlen[v[i]][u[i]]=w[i],tim[u[i]][v[i]]=tim[v[i]][u[i]]=1; else if (w[i]==minlen[u[i]][v[i]]) tim[u[i]][v[i]]++,tim[v[i]][u[i]]++; } for (int i=1; i<=n; i++) { C[i]=dijkstra(i,m+1,dis,from,key); ans+=C[i]; for (int j=1; j<=m; j++) if (!key[j] || tim[u[j]][v[j]]>1 || minlen[u[j]][v[j]]!=w[j]) f[i][j]=C[i]; else f[i][j]=dijkstra(i,j,dis,from,akey); } for (int i=1; i<=m; i++) { ll tmp=0; for (ll j=1; j<=n; j++) tmp+=f[j][i]; sum=max(sum,tmp); } printf("%lld %lld ",ans,sum); } return 0; }