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
题目链接：POJ：http://poj.org/problem?id=1511
HDU：http://acm.hdu.edu.cn/showproblem.php?pid=1535
题目大意：一张有向图，求从点1开始到其他各点的最短路权值和加上从其他各点到点1的最短路权值和
题目分析：这题HDU的数据比较水，给的时间也短，POJ的数据超int，wa了几发，做法就是1到各点直接spfa搞一下，反过来，就把图也反过来建一下，再搞一次spfa就可以了，一套好板子也是很重要的，书上板子动态分配伤不起，其实vector也挺费时，在poj跑了4000ms+，但是比较好写
POJ:
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#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 <NODE> 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 <int> 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 <cstdio>
#include <cstring>
#include <vector>
#include <queue>
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 <NODE> 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 <int> 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);
}
}