标签：rman bin case its special most members 数据规模 determine
Invitation Cards
InputThe input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case begins with a line containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is the number of stops including CCS and Q the number of bus lines. Then there are Q lines, each describing one bus line. Each of the lines contains exactly three numbers - the originating stop, the destination stop and the price. The CCS is designated by number 1. Prices are positive integers the sum of which is smaller than 1000000000. You can also assume it is always possible to get from any stop to any other stop.
OutputFor each case, print one line containing the minimum amount of money to be paid each day by ACM for the travel costs of its volunteers.
Sample Input
2 2 2 1 2 13 2 1 33 4 6 1 2 10 2 1 60 1 3 20 3 4 10 2 4 5 4 1 50
Sample Output
46 210
题意：在有向图中，求一个点到所有点的最短路与所有点到这个点的最短路之和。
思路：正反向建图，两边SPFA。由于数据规模较大，这道题用vector过不了，相比之下前向星存图显得效率更高。以下是前向星建图代码。
#include<stdio.h> #include<string.h> #include<deque> #define MAX 1000005 #define INF 10000000000000000 using namespace std; struct Node{ int v,next,w; }edge[MAX],redge[MAX]; long long dis[MAX],diss[MAX]; int b[MAX],head1[MAX],head2[MAX]; int n,cnt1,cnt2; void Init() { cnt1=0; memset(head1,-1,sizeof(head1)); cnt2=0; memset(head2,-1,sizeof(head2)); } void addEdge(int u,int v,int w) { edge[cnt1].v=v; edge[cnt1].w=w; edge[cnt1].next=head1[u]; head1[u]=cnt1++; } void addrEdge(int u,int v,int w) { redge[cnt2].v=v; redge[cnt2].w=w; redge[cnt2].next=head2[u]; head2[u]=cnt2++; } void spfa(int k) { int i; deque<int> q; for(i=1;i<=n;i++){ dis[i]=INF; diss[i]=INF; } memset(b,0,sizeof(b)); b[k]=1; dis[k]=0; q.push_back(k); while(q.size()){ int u=q.front(); for(i=head1[u];i!=-1;i=edge[i].next){ int v=edge[i].v; int w=edge[i].w; if(dis[v]>dis[u]+w){ dis[v]=dis[u]+w; if(b[v]==0){ b[v]=1; if(dis[v]>dis[u]) q.push_back(v); else q.push_front(v); } } } b[u]=0; q.pop_front(); } b[k]=1; diss[k]=0; q.push_back(k); while(q.size()){ int u=q.front(); for(i=head2[u];i!=-1;i=redge[i].next){ int v=redge[i].v; int w=redge[i].w; if(diss[v]>diss[u]+w){ diss[v]=diss[u]+w; if(b[v]==0){ b[v]=1; if(diss[v]>diss[u]) q.push_back(v); else q.push_front(v); } } } b[u]=0; q.pop_front(); } } int main() { int t,m,u,v,w,i; scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); Init(); for(i=1;i<=m;i++){ scanf("%d%d%d",&u,&v,&w); addEdge(u,v,w); addrEdge(v,u,w); } spfa(1); long long sum=0; for(i=1;i<=n;i++){ sum+=dis[i]+diss[i]; } printf("%lld\n",sum); } return 0; }
HDU - 1535 Invitation Cards 前向星SPFA
标签：rman bin case its special most members 数据规模 determine
原文地址：http://www.cnblogs.com/yzm10/p/7286498.html
