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

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;
}