POJ_1511_Invitation Cards(最短路)
Invitation Cards
Time Limit: 8000MS
Memory Limit: 262144K
Total Submissions: 21615
Accepted: 7089
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
題意：在有向圖中，求1到所有點的最短路之和 + 所有點到1的最短路之和。
分析：很明顯地，利用spfa or dijkstra+heap，順著存邊求一次最短路，然後把邊反向求一次最短路，然後求和即可。如果這個題目用vector存圖的話，那麼很容易卡時間，比如存圖我用兩個臨接表直接存儲就TLE了，後來改成一個臨接表才過的。然而這樣子還是卡時間，還可以繼續優化。題意很清晰，邊的數量是1000000，那麼我們可以直接用結構體數組把邊存下來，這樣子在數組上操作，就不會很費時了。具體見代碼：
題目鏈接：http://poj.org/problem?id=1511
代碼清單：
vector實現（時間復雜度高）：
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int maxn=1000000+5;
const int max_dis=1e9 + 5;
struct Edge{
int to,dis;
Edge(int to,int dis){
this -> to = to;
this -> dis = dis;
}
};
struct edge{
int from,to,dis;
}g[maxn];
int T;
int n,Q;
int a,b,c;
int d[maxn];
bool vis[maxn];
typedef pairP;
vectorgraph[maxn];
void Clear(){
for(int i=1;i<=n;i++){
graph[i].clear();
}
}
void input(){
scanf(%d%d,&n,&Q);
Clear();
for(int i=0;i,greater >q;
while(q.size()) q.pop();
d[1]=0;
q.push(P(0,1));
while(q.size()){
P p=q.top(); q.pop();
int v=p.second;
if(d[v]d[v]+e.dis){
d[e.to]=d[v]+e.dis;
q.push(P(d[e.to],e.to));
}
}
}
ll ret=0;
for(int i=1;i<=n;i++) ret+=d[i];
return ret;
}
ll spfa(){
memset(vis,false,sizeof(vis));
fill(d+1,d+1+n,max_dis);
queueq;
while(!q.empty()) q.pop();
d[1]=0;
vis[1]=true;
q.push(1);
while(!q.empty()){
int p=q.front(); q.pop();
vis[p]=0;
for(int i=0;id[p]+e.dis){
d[e.to]=d[p]+e.dis;
if(!vis[e.to]){
vis[e.to]=true;
q.push(e.to);
}
}
}
}
ll ret=0;
for(int i=1;i<=n;i++) ret+=d[i];
return ret;
}
void dijkstra_solve(){
ll ans=dijkstra();
Clear();
for(int i=0;i
結構體數組實現（時間復雜度較低）：
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int maxn=1000000+5;
const int max_dis=1e9 + 5;
struct Edge{int to,dis,next;}graph[maxn];
struct edge{int from,to,dis;}g[maxn];
int T;
int n,Q;
int num;
int d[maxn];
bool vis[maxn];
int head[maxn];
typedef pairP;
void Clear(){
num=0;
memset(head,-1,sizeof(head));
memset(graph,0,sizeof(graph));
}
void add(int u,int v,int dis){
graph[num].to=v;
graph[num].dis=dis;
graph[num].next=head[u];
head[u]=num++;
}
void input(){
scanf(%d%d,&n,&Q);
Clear();
for(int i=0;i,greater >q;
while(q.size()) q.pop();
d[1]=0;
q.push(P(0,1));
while(q.size()){
P p=q.top(); q.pop();
int v=p.second;
if(d[v]-1;k=graph[k].next){
if(d[graph[k].to]>d[v]+graph[k].dis){
d[graph[k].to]=d[v]+graph[k].dis;
q.push(P(d[graph[k].to],graph[k].to));
}
}
}
ll ret=0;
for(int i=1;i<=n;i++) ret+=d[i];
return ret;
}
ll spfa(){
memset(vis,false,sizeof(vis));
fill(d+1,d+1+n,max_dis);
queueq;
while(!q.empty()) q.pop();
d[1]=0;
vis[1]=true;
q.push(1);
while(!q.empty()){
int v=q.front(); q.pop();
vis[v]=0;
for(int k=head[v];k>-1;k=graph[k].next){
if(d[graph[k].to]>d[v]+graph[k].dis){
d[graph[k].to]=d[v]+graph[k].dis;
if(!vis[graph[k].to]){
vis[graph[k].to]=true;
q.push(graph[k].to);
}
}
}
}
ll ret=0;
for(int i=1;i<=n;i++) ret+=d[i];
return ret;
}
void dijkstra_solve(){
ll ans=dijkstra();
Clear();
for(int i=0;i
