HDU-2066-一个人的旅行

描述

题解

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

/*
*  单源最短路径，Dijkstra算法，邻接矩阵形式，复杂度为O(n^2)
*  求出源beg到所有点的最短路径，传入图的顶点数和邻接矩阵cost[][]
*  返回各点的最短路径lowcost[]，路径pre[]，pre[i]记录beg到i路径上的父节点，pre[beg] = -1
*  可更改路径权类型，但是权值必须为非负，下标0~n
*/
const int MAXN = 1010;
const int INF = 0x3f3f3f3f; //  表示无穷
bool vis[MAXN];
int pre[MAXN];

void Dijkstra(int cost[][MAXN], int lowcost[], int n, int beg)
{
for (int i = 0; i <= n; i++)
{
lowcost[i] = INF;
vis[i] = false;
pre[i] = -1;
}
lowcost[beg] = 0;
for (int j = 0; j <= n; j++)
{
int k = -1;
int min = INF;
for (int i = 0; i <= n; i++)
{
if (!vis[i] && lowcost[i] < min)
{
min = lowcost[i];
k = i;
}
}
if (k == -1)
{
break;
}
vis[k] = true;
for (int i = 0; i <= n; i++)
{
if (!vis[i] && lowcost[k] + cost[k][i] < lowcost[i])
{
lowcost[i] = lowcost[k] + cost[k][i];
pre[i] = k;
}
}
}
}

int T, S, D, num;
int lowcost[MAXN];
int cost[MAXN][MAXN];
int want[MAXN];

int main()
{
int x, y, z;
while (~scanf("%d%d%d", &T, &S, &D))
{
num = 0;
memset(cost, 0x3f, sizeof(cost));

while (T--)
{
scanf("%d%d%d", &x, &y, &z);
num = max(max(num, x), y);
if (z < cost[x][y])
{
cost[x][y] = cost[y][x] = z;
}
}
int minn = INF;
for (int i = 0; i < S; i++)
{
scanf("%d", &x);
cost[0][x] = cost[x][0] = 0;
}
for (int i = 0; i < D; i++)
{
scanf("%d", &want[i]);
}

Dijkstra(cost, lowcost, num, 0);

for (int i = 0; i < D; i++)
{
minn = min(minn, lowcost[want[i]]);
}
printf("%d\n", minn);
}

return 0;
}

参考