Invitation Cards
HDU: Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
POJ: Time Limit: 8000 MS Memory Limit: 262144 K
Problem 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
題目鏈接：POJ：http://poj.org/problem?id=1511
HDU：http://acm.hdu.edu.cn/showproblem.php?pid=1535
題目大意：一張有向圖，求從點1開始到其他各點的最短路權值和加上從其他各點到點1的最短路權值和
題目分析：這題HDU的數據比較水，給的時間也短，POJ的數據超int，wa了幾發，做法就是1到各點直接spfa搞一下，反過來，就把圖也反過來建一下，再搞一次spfa就可以了，一套好板子也是很重要的，書上板子動態分配傷不起，其實vector也挺費時，在poj跑了4000ms+，但是比較好寫
POJ:
#include
#include
#include
#include
#define ll long long
int const MAX = 1e6 + 5;
ll const INF = 0xffffffff;
using namespace std;
struct NODE
{
int v, w;
NODE(int vv, ll ww)
{
v = vv;
w = ww;
}
};
struct EDGE
{
int u, v;
ll w;
}e[MAX];
vector vt[MAX];
int p, q;
ll ans, dist[MAX];
bool vis[MAX];
void SPFA(int v0)
{
memset(vis, false, sizeof(vis));
for(int i = 0; i <= p; i++)
dist[i] = INF;
dist[v0] = 0;
queue Q;
Q.push(v0);
while(!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = false;
int sz = vt[u].size();
for(int i = 0; i < sz; i++)
{
int v = vt[u][i].v;
ll w = vt[u][i].w;
if(dist[v] > dist[u] + w)
{
dist[v] = dist[u] + w;
if(!vis[v])
{
Q.push(v);
vis[v] = true;
}
}
}
}
}
void Clear()
{
for(int i = 1; i <= p; i++)
vt[i].clear();
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
ans = 0;
scanf("%d %d", &p, &q);
for(int i = 0; i < q; i++)
scanf("%d %d %lld", &e[i].u, &e[i].v, &e[i].w);
Clear();
for(int i = 0; i < q; i++)
vt[e[i].u].push_back(NODE(e[i].v, e[i].w));
SPFA(1);
for(int i = 1; i <= p; i++)
ans += dist[i];
Clear();
for(int i = 0; i < q; i++)
vt[e[i].v].push_back(NODE(e[i].u, e[i].w));
SPFA(1);
for(int i = 1; i <= p; i++)
ans += dist[i];
printf("%lld\n", ans);
}
}
HDU:
#include
#include
#include
#include
int const MAX = 1e6 + 5;
int const INF = 0xfffffff;
using namespace std;
struct NODE
{
int v, w;
NODE(int vv, int ww)
{
v = vv;
w = ww;
}
};
struct EDGE
{
int u, v, w;
}e[MAX];
vector vt[MAX];
int p, q, ans, dist[MAX];
bool vis[MAX];
void SPFA(int v0)
{
memset(vis, false, sizeof(vis));
for(int i = 0; i <= p; i++)
dist[i] = INF;
dist[v0] = 0;
queue Q;
Q.push(v0);
while(!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = false;
int sz = vt[u].size();
for(int i = 0; i < sz; i++)
{
int v = vt[u][i].v;
int w = vt[u][i].w;
if(dist[v] > dist[u] + w)
{
dist[v] = dist[u] + w;
if(!vis[v])
{
Q.push(v);
vis[v] = true;
}
}
}
}
}
void Clear()
{
for(int i = 1; i <= p; i++)
vt[i].clear();
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
ans = 0;
scanf("%d %d", &p, &q);
for(int i = 0; i < q; i++)
scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
Clear();
for(int i = 0; i < q; i++)
vt[e[i].u].push_back(NODE(e[i].v, e[i].w));
SPFA(1);
for(int i = 1; i <= p; i++)
ans += dist[i];
Clear();
for(int i = 0; i < q; i++)
vt[e[i].v].push_back(NODE(e[i].u, e[i].w));
SPFA(1);
for(int i = 1; i <= p; i++)
ans += dist[i];
printf("%d\n", ans);
}
}
