题目意思是找在图中与原点距离为L的点的个数,这些点可以在图中的点上,也可以在边上。
先用dijkstra求出原点到各个点的最短距离,然后先扫一遍点,加上距离为L的点,再扫一遍边,分几种情况去判断在边上能有几个符合条件的点(1个或2个),这里写的时候仔细点就可以了。
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include <iostream>
#include<cmath>
using namespace std;
const int INF = 1000000000;
const int maxn = 100000 + 10;
struct Edge {
int from, to, dist, val;
};
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]; // 是否已永久标号
int d[maxn]; // s到各个点的距离
int p[maxn]; // 最短路中的上一条弧
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, int val) {
edges.push_back((Edge){from, to, dist, val});
m = edges.size();
G[from].push_back(m-1);
}
void dijkstra(int s) {
priority_queue<HeapNode> Q;
for(int i = 0; i < n; i++) d[i] = INF;
d[s] = 0;
memset(done, 0, sizeof(done));
Q.push((HeapNode){0, s});
while(!Q.empty()) {
HeapNode x = Q.top(); Q.pop();
int u = x.u;
if(done[u]) continue;
done[u] = true;
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];
Q.push((HeapNode){d[e.to], e.to});
}
}
}
}
vector<int> GetShortestPath(int s, int t) {
vector<int> path;
while(t != s) {
path.push_back(edges[p[t]].val);
t = edges[p[t]].from;
}
reverse(path.begin(), path.end());
return path;
}
};
Dijkstra solver;
struct Node{
int from,to,dis;
};
vector<Node> edg;
int main(){
int n,m,s,l;
while(cin >> n >> m >> s){
s--;
solver.init(n);
edg.clear();
for(int i = 0;i < m;i++){
int v,u,w;
cin >> v >> u >> w;
v--;u--;
edg.push_back(Node{v,u,w});
solver.AddEdge(v,u,w,0);
solver.AddEdge(u,v,w,0);
}
cin >> l;
solver.dijkstra(s);
int ans = 0;
for(int i = 0;i < n;i++){
if(solver.d[i] == l) ans++;
}
for(int i = 0;i < edg.size();i++){
int x = solver.d[edg[i].from];
int y = solver.d[edg[i].to];
if(x > y) swap(x,y);
int dis = edg[i].dis;
if(x + dis < y && x + dis >= l){
if(x < l ) ans++;
if(y < l) ans++;
}
else if(x + dis == y && y != l && x + dis >= l){
if(x < l ) ans++;
if(y < l ) ans++;
}
else if((x+dis > y) && ((x+y+dis)*1.0/2 > l)){
if(x < l ) ans++;
if(y < l ) ans++;
}
else if((x+dis > y) && (fabs(((x+y+dis)*1.0/2) - l) <1e-6) && x < l){
ans++;
}
}
cout << ans << endl;
}
return 0;
}