zoj2008 || poj1511&nbs…

Invitation Cards

Time Limit: 8000MS		Memory Limit: 262144K
Total Submissions: 15659		Accepted: 5089

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
详解见：图论算法
算法：spfa求最短路径

// I'm the
Topcoder
//C
#include
<stdio.h>
#include
<stdlib.h>
#include
<string.h>
#include
<ctype.h>
#include
<math.h>
#include
<time.h>
//C++
#include
<iostream>
#include
<algorithm>
#include
<cstdio>
#include
<cstdlib>
#include
<cmath>
#include
<cstring>
#include
<cctype>
#include
<stack>
#include
<string>
#include
<list>
#include
<queue>
#include
<map>
#include
<vector>
#include
<deque>
#include
<set>
using namespace
std;

//*************************OUTPUT*************************
#ifdef
WIN32
#define INT64
"%I64d"
#define UINT64
"%I64u"
#else
#define INT64
"%lld"
#define UINT64
"%llu"
#endif

//**************************CONSTANT***********************
#define INF
0x3f3f3f3f
#define eps
1e-8
#define PI
acos(-1.)
#define PI2 asin
(1.);
typedef long long
LL;
//typedef __int64 LL;
//codeforces
typedef unsigned int
ui;
typedef unsigned long long
ui64;
#define MP
make_pair
typedef
vector<int> VI;
typedef
pair<int, int>
PII;
#define pb
push_back
#define mp
make_pair

//***************************SENTENCE************************
#define CL(a,b) memset (a, b,
sizeof (a))
#define sqr(a,b) sqrt
((double)(a)*(a) + (double)(b)*(b))
#define sqr3(a,b,c)
sqrt((double)(a)*(a) + (double)(b)*(b) +
(double)(c)*(c))

//****************************FUNCTION************************
template
<typename T> double DIS(T va, T vb) {
return sqr(va.x - vb.x, va.y - vb.y); }
template <class
T> inline T INTEGER_LEN(T v) { int len = 1; while (v
/= 10) ++len; return len; }
template
<typename T> inline T square(T va, T
vb) { return va * va + vb * vb; }

// aply for the memory of the
stack
//#pragma comment (linker,
"/STACK:1024000000,1024000000")
//end

const int maxn =
1000000+100;
struct
node{

int to;

int next;

long long weight;
};
node
edge[maxn],edge1[maxn];//保存边的起点和终点
int n,m;
long long
val;
int
tot,tot1;
int
src;//起点
int
head[maxn],head1[maxn];
int
visit[maxn],visit1[maxn];
long long
dis[maxn],dis1[maxn];

void add(int a,int b,long long
c){

edge[tot].to=b;

edge[tot].weight=c;

edge[tot].next=head[a];

head[a]=tot++;
}

void add1(int a,int b,long long
c){

edge1[tot1].to=b;

edge1[tot1].weight=c;

edge1[tot1].next=head1[a];

head1[a]=tot1++;
}

void
spfa(){

//初始化

for(int
i=1;i<=n;i++){


dis[i]=INF;


visit[i]=0;//访问标记

}

dis[src]=0; visit[src]=1;

int u;

int v;

queue<int>
Q;//优先队列

Q.push(src);

while(!Q.empty()){


u=Q.front();


Q.pop();


visit[u]=0;//必须是0，这题是1也能过不过是错的

//
visit[u]=1;

for(int
i=head[u];i!=-1;i=edge[i].next){



v=edge[i].to;



if(dis[v]>dis[u]+edge[i].weight){




dis[v]=dis[u]+edge[i].weight;



if(!visit[v]){




Q.push(v);




visit[v]=1;



}


}


}

}
}

void
spfa1(){

//初始化

for(int
i=1;i<=n;i++){


dis1[i]=INF; visit1[i]=0;

}

dis1[src]=0;
visit1[src]=1;

int u,v;

queue<int>
Q;

Q.push(src);

while(!Q.empty()){


u=Q.front();


Q.pop();


visit[u]=0;//必须是0，这题是1也能过不过是错的


//visit1[u]=1;

for(int
i=head1[u];i!=-1;i=edge1[i].next){



v=edge1[i].to;



if(dis1[v]>dis1[u]+edge1[i].weight){




dis1[v]=dis1[u]+edge1[i].weight;



if(!visit1[v]){




Q.push(v);




visit1[v]=1;



}


}


}

}
}

int main(){

int a,b,cas;


scanf("%d",&cas);

while(cas--){


scanf("%d%d",&n,&m);


tot=tot1=0;//边的条数

for(int
i=1;i<=n;i++){



head[i]=-1;



head1[i]=-1;


}

for(int
i=1;i<=m;i++){



scanf("%d%d%lld",&a,&b,&val);



add(a,b,val);//正向边



add1(b,a,val);//反向边


}


src=1;//起点（终点）


spfa();


spfa1();

long long
sum=0;

for(int
i=2;i<=n;i++){



sum+=dis[i]+dis1[i];


}


printf("%lld\n",sum);

}

return 0;
}