CF449B Jzzhu and Cities

题目简述

一张无项图有$n$个点，$m$条边。再给你$k$条边，问这些边最少可以保留多少，仍使点$1$到每个店的距离不变。

思路

先将$k$个点上岛（设为最短路），再看剩下的边能否松弛这些点，如果可以就说明这个点与$1$的边可以删除。

code

#include 

using namespace std;

const int N = 1e5 + 10, INF = 2e9;
int cnt;
int head[N];

vector <pair<int ,int > > g[N];

int n, m, k;
bool vis[N];
long long dis[N];
bool book[N];
queue <int> q;

void dij()
{
	while (!q.empty()) 
	{
		int u = q.front();
		q.pop();
		vis[u] = 0;
		for (int i = 0; i < g[u].size(); i++) 
		{
			int v = g[u][i].first;
			int w = g[u][i].second;
			if (dis[v] >= dis[u] + w) 
			{
				dis[v] = dis[u] + w;
				book[v] = 0;
				if (!vis[v])
				{
					vis[v] = 1;
					q.push(v);
				}
			}
		}
	}
}

int main() 
{
	int u, v, w; 
	memset(head, -1, sizeof(head));
	cin >> n >> m >> k;
	for (int i = 1; i <= n; i++) dis[i] = INF;
	for (int i = 1; i <= m; i++) 
	{
		cin >> u >> v >> w;
		g[u].push_back(make_pair(v,w));
		g[v].push_back(make_pair(u,w));
	}
	q.push(1);
	dis[1] = 0;
	vis[1] = 1;
	for(int i = 1; i <= k; i++) 
	{
		cin >> v >> w;
		if (w < dis[v]) 
		{
			dis[v] = w;
			vis[v] = 1;
			book[v] = 1;
			q.push(v);
		}
	}
	dij();
	int ans = 0;
	for(int i = 1; i <= n; i++)
		if(book[i]) ans++;
	cout << k - ans << endl;
	return 0;
}