oyjpArt ACM/ICPC算法程序設計空間

Invitation Cards
Time Limit:3000MS? Memory Limit:65536K
Total Submit:241 Accepted:93

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

Source
Central Europe 1998

這個題目我想就是?？糄ijkstra()的堆寫法吧 中間我采用的逆轉有向邊的寫法 可以把從多點到單源的最短路徑用單源到多點的最短路徑的方法求出。
中間出了一個很隱蔽的錯誤 在一個i, j的雙層循環(huán)中 內(nèi)層循環(huán)寫成了i++...結果調(diào)了很久。。。
總算過了 發(fā)現(xiàn)STL占內(nèi)存是直接寫鄰接表的2倍左右，這也印證了vector的擴張方式。

??1 Solution:
??2 // by?Optimistic
??3 #include? < stdio.h >
??4 #include? < string .h >
??5 #include? < vector >
??6 using ? namespace ?std;?
??7
??8 const ? int ?MAXINT? = ? 200000000 ;
??9 // const?double?INF?=?10e100;
?10 // const?double?EPS?=?10e-6;?
?11
?12 const ? int ?N? = ? 1000010 ;
?13 int ?nv,?ne;
?14 typedef? struct { int ?jj,?w;} Vtx;
?15 vector < Vtx > ?adj[N];
?16 vector < Vtx > ?adj2[N];
?17 typedef? struct { int ?k,?no;} hNode;
?18 int ?ntc,?hs;
?19 hNode?h[N];?
?20
?21 bool ? operator ? < ?( const ?hNode & ?a,? const ?hNode & ?b)?
?22 {
?23 ? return ?a.k? < ?b.k;
?24 } ?
?25
?26 void ?push(hNode?t)
?27 {
?28 ? int ?i? = ? ++ hs;
?29 ? while (i? > ? 1 ? && ?t? < ?h[i >> 1 ])? {
?30 ??h[i]? = ?h[i >> 1 ];
?31 ??i? >>= ? 1 ;
?32 ?}
?33 ?h[i]? = ?t;
?34 } ?
?35
?36 void ?pop()
?37 {
?38 ?hs -- ;
?39 ? int ?i? = ? 1 ,?ic? = ? 2 ;
?40 ? while (ic? <= ?hs)? {
?41 ?? if (ic + 1 ? <= ?hs? && ?h[ic + 1 ]? < ?h[ic])?ic ++ ;
?42 ?? if (h[hs + 1 ]? < ?h[ic])? break ;
?43 ??h[i]? = ?h[ic];
?44 ??i? = ?ic;
?45 ??ic? <<= ? 1 ;
?46 ?}
?47 ?h[i]? = ?h[hs + 1 ];
?48 } ?
?49
?50 int ?Dijkstra()
?51 {
?52 ?hs? = ? 0 ;
?53 ? int ?i;
?54 ? int ? * ?dist? = ? new ? int [nv];
?55 ? for (i? = ? 0 ;?i? < ?nv;?i ++ )?dist[i]? = ?MAXINT;
?56 ?hNode?now;
?57 ?now.k? = ? 0 ;?now.no? = ? 0 ;
?58 ?push(now);
?59 ? while ( 1 )
?60 ? {
?61 ?? while (hs? > ? 0 ? && ?h[ 1 ].k? > ?dist[h[ 1 ].no])?
?62 ???pop();
?63 ?? if (hs? == ? 0 )? break ;
?64 ??now? = ?h[ 1 ];
?65 ??pop();
?66 ?? int ?u? = ?now.no;
?67 ??dist[u]? = ?now.k;
?68 ?? for (i? = ? 0 ;?i? < ?adj[u].size();?i ++ )
?69 ?? {
?70 ??? int ?v? = ?adj[u][i].jj;
?71 ??? int ?w? = ?adj[u][i].w;
?72 ??? if (dist[v]? > ?dist[u]? + ?w)
?73 ??? {
?74 ????now.k? = ?dist[u]? + ?w;
?75 ????now.no? = ?v;
?76 ????push(now);
?77 ????dist[v]? = ?dist[u]? + ?w;
?78 ???}
?79 ??}
?80 ?}
?81 ? int ?ans? = ? 0 ;
?82 ? for (i? = ? 0 ;?i? < ?nv;?i ++ )
?83 ??ans? += ?dist[i];
?84 ? return ?ans;
?85 } ?
?86
?87 void ?init()
?88 {
?89 ? int ?i,?u,?v,?w;
?90 ?Vtx?x;
?91 ? // initiation
?92 ? for (i? = ? 0 ;?i? < ?nv;?i ++ )? {
?93 ??adj2[i].clear();
?94 ??adj[i].clear();
?95 ?}
?96 ? // input
?97 ?scanf( " %d?%d " ,? & nv,? & ne);
?98 ? for (i? = ? 0 ;?i? < ?ne;?i ++ )? {
?99 ??scanf( " %d?%d?%d " ,? & u,? & v,? & w);
100 ??u -- ;?v -- ;?
101 ??x.jj? = ?v;?x.w? = ?w;
102 ??adj[u].push_back(x);
103 ?}
104 ? // pretreatment
105 } ?
106
107 void ?Reverse()
108 {
109 ? int ?i,?j;
110 ?Vtx?x;
111 ? for (i? = ? 0 ;?i? < ?nv;?i ++ )
112 ??adj2[i].clear();
113 ? for (i? = ? 0 ;?i? < ?nv;?i ++ )
114 ? {
115 ?? for (j? = ? 0 ;?j? < ?adj[i].size();?j ++ )
116 ?? {
117 ???x.jj? = ?i;
118 ???x.w? = ?adj[i][j].w;
119 ???adj2[adj[i][j].jj].push_back(x);
120 ??}
121 ?}
122 ? for (i? = ? 0 ;?i? < ?nv;?i ++ )
123 ? {
124 ??adj[i].clear();
125 ?? for (j? = ? 0 ;?j < adj2[i].size();?j ++ )
126 ???adj[i].push_back(adj2[i][j]);
127 ?}
128 } ?
129
130 void ?work()
131 {
132 ? int ?x? = ?Dijkstra();
133 ?Reverse();
134 ?x? += ?Dijkstra();
135 ?printf( " %d\n " ,?x);
136 } ?
137
138 int ?main()
139 {
140 // ?freopen("t.in",?"r",?stdin);
141 ?scanf( " %d " ,? & ntc);
142 ? while (ntc -- )
143 ? {
144 ??init();
145 ??work();
146 ?}
147 ? return ? 0 ;
148 } ?
149