题意:给定一个图,可以在边和点上设立发射井,其中发射井到首都的距离必须等于L,问可以建多少个发射井
思路:建图然后跑最短路,如果发射井在边上的话,画个图可以知道限制的条件(利用三角不等式),如果发射井在点上的话只需要判是否等于L即可,见代码
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 200005
#define LL long long
int cas=1,T;
int n,m,w;
const int INF = 0x3f3f3f3f ;
struct Edge
{
int from,to,dist;
Edge(){}
Edge(int u,int v,int d):from(u),to(v),dist(d){}
};
vector<Edge>edges;
vector<int> G[maxn];
int d[maxn];
int inq[maxn];
int cnt[maxn];
struct Bellman_ford
{
void init()
{
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));
int mm = edges.size();
G[from].push_back(mm-1);
}
bool bellman_ford(int s)
{
queue<int>q;
memset(inq,0,sizeof(inq));
memset(cnt,0,sizeof(cnt));
for (int i =0;i<=n;i++)
d[i]=INF;
d[s]=0;
inq[s]=1;
q.push(s);
while (!q.empty())
{
int u = q.front();q.pop();
inq[u] = false;
for (int i =0;i<G[u].size();i++)
{
Edge &e = edges[G[u][i]];
if (d[u]<INF && d[e.to]>d[u]+e.dist)
{
d[e.to] = d[u] + e.dist;
if (!inq[e.to])
{
q.push(e.to);
inq[e.to]=1;
if (++cnt[e.to]>n)
return false;
}
}
}
}
return true;
}
};
int main()
{
//freopen("in","r",stdin);
int s;
scanf("%d%d%d",&n,&m,&s);
Bellman_ford bf;
bf.init();
for (int i = 0;i<m;i++)
{
int u,v,t;
scanf("%d%d%d",&u,&v,&t);
bf.AddEdge(u,v,t);
bf.AddEdge(v,u,t);
}
int l;
scanf("%d",&l);
bf.bellman_ford(s);
int ans = 0;
for (int i = 1;i<=n;i++)
{
for (int j = 0;j<G[i].size();j++)
{
Edge e = edges[G[i][j]];
if (d[e.from] < l && l-d[e.from]<e.dist && (d[e.to]+e.dist-(l-d[e.from])> l))
ans++;
if (d[e.to] < l && l-d[e.to]<e.dist && (d[e.from]+e.dist-(l-d[e.to])> l))
ans++;
if (d[e.from] < l && d[e.to]<l && (d[e.from]+d[e.to]+e.dist==2*l))
ans++;
}
}
ans = ans /2; //对于每一条边计算了两次,所以除二
for (int i = 1;i<=n;i++)
if (d[i]==l)
ans++;
printf("%d\n",ans);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}