Uva 10917 Walk Through the Forest-优快云博客

本文链接：https://blog.youkuaiyun.com/MakingMaker/article/details/8316041

本文详细介绍了一种使用Dijkstra算法求解最短路径问题的方法，并通过一个具体的实现案例展示了如何运用堆优化来提高算法效率。此外，还介绍了如何利用该算法进行树形结构的遍历并计算节点间的距离。

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完全没考虑原理瞬间敲出来然后WA了。。。。。坑爹啊

#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;

const int MAXN = 1007;
const ll INF = 1LL<<56;
struct Dijkstra {
	struct HeapNode {
		ll d, u;
		HeapNode(ll d, int u):d(d), u(u) {}
		bool operator < (const HeapNode &hn) const {
			return d > hn.d;
		}
	};
	vector<int> G[MAXN];
	vector<ll> W[MAXN];
	bool done[MAXN];
	int n;
	void init(int n) {
		this->n = n;
		int i;
		for(i = 1; i <= n; i++) {
			G[i].clear();
			W[i].clear();
		}
	}
	void add(int u, int v, ll dist) {
		G[u].push_back(v);
		W[u].push_back(dist);
	}
	void solve(int s, ll *d) {
		memset(done, 0, sizeof(done));
		int i, u, v, sz;
		for(i = 1; i <= n; i++) d[i] = INF;
		priority_queue<HeapNode> pq;
		d[s] = 0;
		pq.push(HeapNode(0, s));
		while(!pq.empty()) {
			u = pq.top().u; pq.pop();
			if(done[u]) continue;
			done[u] = true;
			sz = G[u].size();
			for(i = 0; i < sz; i++) {
				v = G[u][i];
				if(d[v] > d[u] + W[u][i]) {
					d[v] = d[u] + W[u][i];
					pq.push(HeapNode(d[v], v));
				}
			}
		}
	}
};
Dijkstra dijk;
ll d[MAXN], dp[MAXN];
vector<int> G[MAXN];

ll getdp(int u) {
	if(dp[u] != -1) return dp[u];
	dp[u] = 0;
	int i, sz = G[u].size();
	for(i = 0; i < sz; i++)
		dp[u] += getdp(G[u][i]);
	return dp[u];
}

int main() {
	int n, m;
	while(~scanf("%d", &n), n) {
		scanf("%d", &m);
		int i, j, sz, u, v;
		ll w;
		dijk.init(n);
		for(i = 1; i <= n; i++) G[i].clear();
		for(i = 1; i <= m; i++) {
			scanf("%d%d%lld", &u, &v, &w);
			dijk.add(u, v, w);
			dijk.add(v, u, w);
		}
		dijk.solve(2, d);
		for(i = 1; i <= n; i++) {
			sz = dijk.G[i].size();
			for(j = 0; j < sz; j++) {
				v = dijk.G[i][j];
				if(d[i] > d[v]) G[i].push_back(v);
			}
		}
		memset(dp, -1, sizeof(dp));
		dp[2] = 1;
		printf("%lld\n", getdp(1));
	}
	return 0;
}