HDU Travel for 20,661 people

http://acm.hdu.edu.cn/showproblem.php?pid=2066

Analysis:

The shortest time for the city where the grass wants to go is the shortest distance from the home->a (neighboring city)->b (any one of several).

Since the end is from a starting point to an end point, careful analysis, it is not difficult to conclude that this is a single source shortest path problem (is a starting point, an end point).

So you can use Dijkstra.

The code is as follows:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#define MAXX 999999999
using namespace Std;
int T, S, D, N,u;
int map[1005][1005];
int star[1005];
int terminal[1005];
int _distance[1005];
int vis[1005];
void Djstl ()
{
memset (Vis, 0, sizeof (VIS));
for (int i = 0;i <= n;++i)
{
_distance[i] = Map[0][i];

This represents the distance from all points to 0 points; using _distance[i] to represent
}
Vis[0] = 1;

Mark 0 points have passed-------------------------------------
for (int i = 1;i <= n-1;++i)
{
int mins = Maxx;
for (int j = 1;j <= n;++j)
{
if (!vis[j] && mins > _distance[j])
{
mins = _distance[j];
U = j;
}
}
if (Vis[u])
Continue

The point closest to I is not tested, and goes directly to the next loop. (Note the position of the U definition)
Vis[u] = 1;
for (int j = 1;j <= n;++j)
{
if (map[u][j]! = Maxx)
if (Map[u][j] + _distance[u] < _distance[j])
_DISTANCE[J] = _distance[u] + map[u][j];

Relaxation operation, distance to U + distance from u to J, more than existing line to J, shorter, update
}
}
}
int main ()
{
while (scanf ("%d%d%d", &t, &s, &d) = = 3)
{
for (int i = 0;i < 1005;++i)
{
for (int j = 0;j < 1005;++j)
MAP[I][J] = Maxx;
Map[i][i] = 0;
}

Using adjacency matrix to save the graph--------------------------------------
int A, B, Tim;
n = 0;
for (int i = 0;i < T;++i)
{
scanf ("%d%d%d", &a, &b, &tim);
MAP[A][B] = Map[b][a] = Map[a][b]>tim? TIM:MAP[A][B];

Save the shortest path between two cities-----------------------------
n = max (max (A, n), b);

Save up to a few cities (save time)-------------------------
}
for (int i = 0;i < S;++i)
{
scanf ("%d", &star[i]);
Map[0][star[i]] = map[star[i]][0] = 0;

Equivalent to having multiple starting points---------------------------------------
}
Djstl ();
int mins = Maxx;
for (int i = 0;i < D;++i)
{
scanf ("%d", &terminal[i]);
if (_distance[terminal[i]] < mins)
mins = _distance[terminal[i]];

Find the shortest distance you can---------------------------------------
}
cout << mins << Endl;
}
}