• Destroying Roads


    题目链接

    (Destroying)

    分析

    又是变形了的最短路
    我们可以考虑哪些道路必须被保留
    然后枚举两个起点到终点重复的道路
    考虑公合法用这些道路就可以了

    (Code)

    #include<cstdio>
    #include<queue>
    #include<cstring>
    #include<iostream> 
    using namespace std;
    
    const int N = 3005;
    int n , m , s1 , t1 , l1 , s2 , t2 , l2;
    int tot , h[N] , dis[N][N] , vis[N];
    struct edge{int to , nxt;}e[2 * N];
    struct node{
    	int id , d;
    	bool operator < (node c) const {return d > c.d;}
    };
    priority_queue<node> Q;
    
    inline void add(int x , int y){e[++tot] = edge{y , h[x]} , h[x] = tot;}
    
    void dijkstra(int s)
    {
    	while (!Q.empty()) Q.pop();
    	memset(vis , 0 , sizeof vis);
    	Q.push(node{s , dis[s][s] = 0});
    	while (!Q.empty())
    	{
    		node now = Q.top(); Q.pop();
    		int u = now.id;
    		if (vis[u]) continue;
    		vis[u] = 1;
    		for(register int i = h[u]; i; i = e[i].nxt)
    		{
    			int v = e[i].to;
    			if (dis[s][u] + 1 < dis[s][v])
    			{
    				dis[s][v] = dis[s][u] + 1;
    				Q.push(node{v , dis[s][v]});
    			}
    		}
    	}
    }
    
    int main()
    {
    	scanf("%d%d" , &n , &m);
    	int x , y;
    	for(register int i = 1; i <= m; i++) 
    		scanf("%d%d" , &x , &y) , add(x , y) , add(y , x);
    	scanf("%d%d%d%d%d%d" , &s1 , &t1 , &l1 , &s2 , &t2 , &l2);
    	memset(dis , 0x3f3f3f3f , sizeof dis);
    	for(register int i = 1; i <= n; i++) dijkstra(i);
    	if (dis[s1][t1] > l1 || dis[s2][t2] > l2)
    	{
    		printf("-1");
    		return 0;
    	}
    	int ans = dis[s1][t1] + dis[s2][t2];
    	for(register int i = 1; i <= n; i++)
    		for(register int j = 1; j <= n; j++)
    		{
    			x = dis[s1][i] + dis[i][j] + dis[j][t1];
    			y = dis[s2][i] + dis[i][j] + dis[j][t2];
    			if (x <= l1 && y <= l2)
    				ans = min(ans , x + y - dis[i][j]);
    			y = dis[t2][i] + dis[i][j] + dis[j][s2];
    			if (x <= l1 && y <= l2)
    				ans = min(ans , x + y - dis[i][j]);
    		}
    	printf("%d" , m - ans);
    }
    
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  • 原文地址:https://www.cnblogs.com/leiyuanze/p/13825945.html
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