【最短路】POJ 1502 MPI Maelstrom

POJ 1502 MPI Maelstrom

题意：给一个无向图，求从结点1到其他结点的最短路中的长度最大值

思路：裸的Dijkstra……记得先cin>>n再初始化，居然在这个弱智错误上卡了10分钟。atoi()在数字与字符混合输入这种情况好用，用来将一个字符串转为整型。

struct node {
	LL d;
	int u;
	bool operator < (const node& k) const {
		return d > k.d;
	}
};

struct Edge {
	int from, to;
	LL dis;
	Edge(int u,int v,LL d):from(u),to(v),dis(d){}
};

int n, m;
int p[maxn];
vector<Edge> Edges;
vector<int> G[maxn];
bool done[maxn];
LL d[maxn];

void solve(){
	cin >> n;
	memset(done, false, sizeof(done));
	memset(p, 0, sizeof(p));
	memset(d, 0, sizeof(d));
	Edges.clear();
	for (int i = 0; i <= n; i++) G[i].clear();
	for (int i = 1; i <= n; i++) { d[i] = (i == 1 ? 0 : INF);}
	Edges.push_back(Edge(0, 0, 0));
	int num = 0;

	for (int i = 2; i <= n; i++) {
		for (int j = 1; j <= i - 1; j++) {
			char ch[9];
			cin >> ch;
			LL d;
			if (ch[0] != 'x')
			{
				d = atoi(ch);
				Edges.push_back(Edge(i, j, d));
				G[i].push_back(++num);
				Edges.push_back(Edge(j, i, d));
				G[j].push_back(++num);
			}
		}
	}
	priority_queue<node> Q;
	Q.push(node{ 0, 1 });
	while (!Q.empty()) {
		node x = Q.top(); Q.pop();
		int u = x.u;
		if (done[u]) continue;
		for (int i = 0; i < G[u].size(); i++) {
			Edge& e = Edges[G[u][i]];
			if (d[u] + e.dis < d[e.to]) {
				d[e.to] = d[u] + e.dis;
				p[e.to] = G[u][i];
				Q.push(node{ d[e.to],e.to });
			}
		}
		done[u] = true;
	}
	
	LL ans = -1;
	for (int i = 2; i <= n; i++) ans = max(ans, d[i]);
	cout << ans;
}