Description
In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards
with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation
to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.
The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait
until the next full half an hour, e.g. X:00 or X:30, where ‘X‘ denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting
and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.
All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program
that helps ACM to minimize the amount of money to pay every day for the transport of their employees.
Input
The 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.
Output
For 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
题解：求1到其他点和其他点的和的最小值，正向建图和逆向建图后两次都求1到其他点的最小值之和就行。但是数据较大，用vector超时，需要自己建立邻接表。
djistra（堆优化 + 动态建图） 7000ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int INF = 1000000008;
struct Node
{
int to;
int cost;
Node(int a,int b)
{
to = a;
cost = b;
}
bool operator< (Node t) const
{
return cost > t.cost;
}
};
struct Node1
{
int from;
int to;
Node1* next;
int cost;
Node1(){
}
Node1(int a,int b,int c,Node1* t = NULL)
{
from = a;
to = b;
cost = c;
next = t;
}
};
Node1 vec[2][1000006];
bool visited[1000006];
int d[1000006];
long long ans;
void prim(int n,int flag)
{
memset(visited,false,sizeof(visited));
for(int i = 1;i <= n;i++)
{
d[i] = INF;
}
d[1] = 0;
priority_queue<Node> q;
q.push(Node(1,0));
while(!q.empty())
{
Node p = q.top();
q.pop();
if(visited[p.to])
{
continue;
}
visited[p.to] = true;
ans += p.cost;
for(Node1* pp = vec[flag][p.to].next;pp != NULL;pp = pp->next)
{
int t = pp->to;
if(!visited[t] && d[t] > d[p.to] + pp->cost)
{
d[t] = d[p.to] + pp->cost;
q.push(Node(t,d[t]));
}
}
}
}
int main()
{
int ncase;
cin>>ncase;
while(ncase--)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;i++)
{
vec[0][i].next = NULL;
vec[1][i].next = NULL;
}
for(int i = 0;i < m;i++)
{
int u,v,c;
scanf("%d%d%d",&u,&v,&c);
Node1* p = new Node1(u,v,c);
p->next = vec[0][u].next;
vec[0][u].next = p;
p = new Node1(v,u,c);
p->next = vec[1][v].next;
vec[1][v].next = p;
}
ans = 0;
prim(n,0);
prim(n,1);
printf("%lld\n",ans);
}
return 0;
}
djistra（堆优化 + 数组邻接表） = 2000ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int INF = 1000000008;
struct Node
{
int from;
int to;
int cost;
Node(){
}
Node(int a,int b,int c)
{
from = a;
to = b;
cost = c;
}
bool operator< (Node t) const
{
return cost > t.cost;
}
};
Node vec[2][1000006];
bool visited[1000006];
int d[1000006];
int first[2][1000006];
int next[2][1000006];
long long ans;
void prim(int n,int flag)
{
memset(visited,false,sizeof(visited));
for(int i = 1;i <= n;i++)
{
d[i] = INF;
}
d[1] = 0;
priority_queue<Node> q;
q.push(Node(1,1,0));
while(!q.empty())
{
Node p = q.top();
q.pop();
if(visited[p.to])
{
continue;
}
visited[p.to] = true;
ans += p.cost;
for(int i = first[flag][p.to];i != -1;i = next[flag][i])
{
int t = vec[flag][i].to;
if(!visited[t] && d[t] > d[p.to] + vec[flag][i].cost)
{
d[t] = d[p.to] + vec[flag][i].cost;
q.push(Node(p.to,t,d[t]));
}
}
}
}
int main()
{
int ncase;
cin>>ncase;
while(ncase--)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;i++)
{
first[0][i] = -1;
first[1][i] = -1;
}
for(int i = 0;i < m;i++)
{
int u,v,c;
scanf("%d%d%d",&u,&v,&c);
vec[0][i].from = u;
vec[0][i].to = v;
vec[0][i].cost = c;
next[0][i] = first[0][u];
first[0][u] = i;
vec[1][i].from = v;
vec[1][i].to = u;
vec[1][i].cost = c;
next[1][i] = first[1][v];
first[1][v] = i;
}
ans = 0;
prim(n,0);
prim(n,1);
printf("%lld\n",ans);
}
return 0;
}
SPFA（动态建图）：7000ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int INF = 1000000008;
struct Node
{
int to;
int cost;
Node(int a,int b)
{
to = a;
cost = b;
}
bool operator< (Node t) const
{
return cost < t.cost;
}
};
struct Node1
{
int from;
int to;
Node1* next;
int cost;
Node1(){
}
Node1(int a,int b,int c,Node1* t = NULL)
{
from = a;
to = b;
cost = c;
next = t;
}
};
Node1 vec[2][1000006];
bool visited[1000006];
int d[1000006];
long long ans;
void spfa(int n,int flag)
{
for(int i = 1;i <= n;i++)
{
d[i] = INF;
}
d[1] = 0;
queue<int> q;
q.push(1);
visited[1] = true;
while(!q.empty())
{
int x = q.front();
q.pop();
visited[x] = false;
for(Node1* p = vec[flag][x].next;p != NULL;p = p->next)
{
int t = p->to;
//cout<<p->cost<<" ";
if(d[t] > d[x] + p->cost)
{
d[t] = d[x] + p->cost;
if(!visited[t])
{
visited[t] = true;
q.push(t);
}
}
}
}
}
int main()
{
int ncase;
cin>>ncase;
while(ncase--)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;i++)
{
vec[0][i].next = NULL;
vec[1][i].next = NULL;
}
for(int i = 0;i < m;i++)
{
int u,v,c;
scanf("%d%d%d",&u,&v,&c);
Node1* p = new Node1(u,v,c);
p->next = vec[0][u].next;
vec[0][u].next = p;
p = new Node1(v,u,c);
p->next = vec[1][v].next;
vec[1][v].next = p;
}
ans = 0;
spfa(n,0);
for(int i = 2;i <= n;i++)
{
ans += d[i];
}
spfa(n,1);
for(int i = 2;i <= n;i++)
{
ans += d[i];
}
printf("%lld\n",ans);
}
return 0;
}
SPFA（数组建图） 1750ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int INF = 1000000008;
struct Node
{
int to;
int cost;
Node(){
}
Node(int a,int b)
{
to = a;
cost = b;
}
bool operator< (Node t) const
{
return cost < t.cost;
}
};
Node vec[2][1000006];
bool visited[1000006];
int d[1000006];
int first[2][1000006];
int next[2][1000006];
long long ans;
void spfa(int n,int flag)
{
for(int i = 1;i <= n;i++)
{
d[i] = INF;
}
d[1] = 0;
queue<int> q;
q.push(1);
visited[1] = true;
while(!q.empty())
{
int x = q.front();
q.pop();
visited[x] = false;
for(int i = first[flag][x];i != -1;i = next[flag][i])
{
int t = vec[flag][i].to;
if(d[t] > d[x] + vec[flag][i].cost)
{
d[t] = d[x] + vec[flag][i].cost;
if(!visited[t])
{
visited[t] = true;
q.push(t);
}
}
}
}
}
int main()
{
int ncase;
cin>>ncase;
while(ncase--)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i = 1;i <= n;i++)
{
first[0][i] = -1; //编号为i的邻接点（该条边的下标）初始化为-1，表示该点没邻接点
first[1][i] = -1;
}
for(int i = 0;i < m;i++)
{
int u,v,c;
scanf("%d%d%d",&u,&v,&c);
vec[0][i].to = v;
vec[0][i].cost = c;
next[0][i] = first[0][u]; //u的新的邻接点取代了u以前的邻接点，新的邻接点保存以前邻接点的位置
first[0][u] = i; //u的第一个邻接点被新的取代，输出u的邻接点时和输入刚好相反
vec[1][i].to = u;
vec[1][i].cost = c;
next[1][i] = first[1][v]; //u的新的邻接点取代了u以前的邻接点，新的邻接点保存以前邻接点的位置
first[1][v] = i; //u的第一个邻接点被新的取代，输出u的邻接点时和输入刚好相反
}
ans = 0;
spfa(n,0);
for(int i = 2;i <= n;i++)
{
ans += d[i];
}
spfa(n,1);
for(int i = 2;i <= n;i++)
{
ans += d[i];
}
printf("%lld\n",ans);
}
return 0;
}
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