kuangbin最短路专题Silver Cow Party (POJ - 3268)

kuangbin最短路专题4

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题面

思路

老方法dijkstra。单向图。正向反向的图都建一遍。正向的图跑奶牛去参加聚会的最短路，反向的图跑回家的最短路。不建反向图就得跑n遍dij

代码

#include <bits/stdc++.h>//poj不能用万能头，这里是为了看起来简洁点
typedef long long ll;
#define mP make_pair
#define pii pair<int, int> 
using namespace std;

int n, m, k;
bool vis[1005];
int dis[1005][1005];
struct lxm
{
	int to, cost, nxt;
}edge[100005], edge1[1000005];
int head[1005], cnt = 0, cnt1 = 0, head1[1005];
int ans[1005], ans1[1005];//这三行另开一组数组存反向边
void addedge(int u, int v, int c)
{
	edge[++cnt].to = v;
	edge[cnt].cost = c;
	edge[cnt].nxt = head[u];
	head[u] = cnt;
	edge1[++cnt1].to = u;
	edge1[cnt1].cost = c;
	edge1[cnt1].nxt = head1[v];
	head1[v] = cnt1;
}
void dij(int s)//老dij了
{
	ans[s] = ans1[s] = 0;
	priority_queue<pii, vector<pii>, greater<pii> >Q;
	Q.push(mP(0, s));
	while (Q.size())
	{
		pii tmp = Q.top(); Q.pop();
		int u = tmp.second;
		if (vis[u])continue;
		vis[u] = 1;
		for (int i = head[u]; i; i = edge[i].nxt)
		{
			if (ans[edge[i].to] > ans[u] + edge[i].cost)
			{
				ans[edge[i].to] = ans[u] + edge[i].cost;
				Q.push(mP(ans[edge[i].to], edge[i].to));
			}
		}
	}
	memset(vis, 0, sizeof(vis));//记得这一步
	Q.push(mP(0, s));//两次dij完全一样，只是改个数组名
	while (Q.size())
	{
		pii tmp = Q.top(); Q.pop();
		int u = tmp.second;
		if (vis[u])continue;
		vis[u] = 1;
		for (int i = head1[u]; i; i = edge1[i].nxt)
		{
			if (ans1[edge1[i].to] > ans1[u] + edge1[i].cost)
			{
				ans1[edge1[i].to] = ans1[u] + edge1[i].cost;
				Q.push(mP(ans1[edge1[i].to], edge1[i].to));
			}
		}
	}
}
void solve()
{
	memset(vis, 0, sizeof(vis));
	memset(ans, 0x3f3f3f3f, sizeof(ans));
	memset(ans1, 0x3f3f3f3f, sizeof(ans1));
	cin >> n >> m >> k;
	for (int i = 1; i <= m; i++)
	{
		int x, y, c;
		cin >> x >> y >> c;
		addedge(x, y, c);
	}
	dij(k);
	int mans = 0;
	for (int i = 1; i <= n; i++)
	{
		if (i == k)continue;
//这里一定要continue，因为ans[k]+ans1[k]==0，要是不跳，mans又回到0了
		mans = max(mans, ans[i] + ans1[i]);
	}
	cout << mans << endl;
}
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	
	//int numoftest; cin >> numoftest;
	//for (register int siven = 1; siven <= numoftest; ++siven)

	solve();
	return 0;
}

/*

*/