UVA10917 A Walk Through the Forest

微微发亮的传送门

思路比较简单，就是先求出来所有点到家的最短路，这个非常好求了，把家当做起点，dijkstra还是spfa都可以搞定，假设用d数组来保存最短路，然后对于某一个点A，如果存在一个点B，使得d[A] > d[B]，就可以从A走到B，也就是说可以建立一条从A到B的有向边，这样的话，我们就能得到一个新图，只需要求从每一个点到终点有多少种方法即可，动态规划完全可以求出来。

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int INF = 0x7fffffff;const int maxn = 1000 + 10;struct Edge{int from, to, dist;};struct HeapNode{int d, u;bool operator < (const HeapNode& rhs) const {return d > rhs.d;}};struct Dijkstra{int n, m;vector<Edge> edges;vector<int> G[maxn];bool done[maxn];int d[maxn], p[maxn];void init(int n){this -> n = n;for (int i = 0; i < n; i++)G[i].clear();edges.clear();}void AddEdge(int from, int to , int dist){edges.push_back((Edge){from, to, dist});m = edges.size();G[from].push_back(m - 1);}void dijkstra(int s){priority_queue<HeapNode> que;for (int i = 0; i < n; i++)d[i] = INF;d[s] = 0;memset(done, 0, sizeof(done));que.push((HeapNode){0, s});while(!que.empty()){HeapNode x = que.top(); que.pop();int u = x.u;if (done[u]) continue;done[u] = 1;for (int i = 0; i < G[u].size(); i++){Edge& e = edges[G[u][i]];if (d[e.to] > d[u] + e.dist){d[e.to] = d[u] + e.dist;p[e.to] = G[u][i];que.push((HeapNode){d[e.to], e.to});}}}}};Dijkstra solver;int n, m, d[maxn];int dp(int u){if (u == 1) return 1;int& ans = d[u];if (ans >= 0) return ans;ans = 0;for (int i = 0; i < solver.G[u].size(); i++){int v = solver.edges[solver.G[u][i]].to;if (solver.d[v] < solver.d[u]) ans += dp(v);}return ans;}int main(){//freopen("in.txt", "r", stdin);while(~scanf("%d", &n)){if (!n) break;scanf("%d", &m);solver.init(n);for (int i = 0; i < m; i++){int a, b, c;scanf("%d%d%d", &a, &b, &c);a -= 1; b -= 1;solver.AddEdge(a, b, c);solver.AddEdge(b, a, c);}solver.dijkstra(1);memset(d, -1, sizeof(d));printf("%d\n", dp(0));}return 0;}