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;
}