uva 10917 - Walk Through the Forest（最短路+记忆化）

来源：互联网发布：剑雨江湖仙器进阶数据编辑：程序博客网时间：2024/06/06 11:05

题目链接：uva 10917 - Walk Through the Forest

做一遍最短路，记录下所有点到终点的距离，然后重新建图，如果d[u] < d[v]那么就在u和v之间建立一条有向边，最后记忆化搜索。

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;typedef pair<int,int> pii;typedef long long ll;const int maxn = 1005;const int inf = 0x3f3f3f3f;int N, M, D[maxn], P[maxn], V[maxn], C[maxn];ll dp[maxn];vector<int> T[maxn];vector<pii> G[maxn];struct State {int dis, idx;State (int dis = 0, int idx = 0):dis(dis), idx(idx) {}bool operator < (const State& u) const { return dis > u.dis; }};void init () {for (int i = 1; i <= N; i++) G[i].clear();int u, v, w;scanf("%d", &M);while (M--) {scanf("%d%d%d", &u, &v, &w);G[u].push_back(make_pair(v, w));G[v].push_back(make_pair(u, w));}}void dijkstra (int s, int e) {memset(V, 0, sizeof(V));memset(D, inf, sizeof(D));D[s] = 0;priority_queue<State> Q;Q.push(State(D[s], s));while (!Q.empty()) {int u = Q.top().idx;Q.pop();if (V[u]) continue;V[u] = 1;for (int i = 0; i < G[u].size(); i++) {int v = G[u][i].first, w = G[u][i].second;if (D[v] > D[u] + w) {D[v] = D[u] + w;Q.push(State(D[v], v));}}}}ll dfs(int u) {ll& ans = dp[u];if (ans != -1) return ans;if (u == 2) return ans = 1;ans = 0;for (int i = 0; i < T[u].size(); i++)ans += dfs(T[u][i]);return ans;}ll solve () {dijkstra(2, 1);memset(dp, -1, sizeof(dp));for (int i = 1; i <= N; i++) {T[i].clear();for (int j = 0; j < G[i].size(); j++) {int v = G[i][j].first, w = G[i][j].second;if (D[i] > D[v])T[i].push_back(v);}}return dfs(1);}int main () {while (scanf("%d", &N) == 1 && N) {init ();printf("%lld\n", solve());}return 0;}

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