zoukankan      html  css  js  c++  java
  • 【洛谷 P2483】 【模板】k短路([SDOI2010]魔法猪学院)(A*)

    题目链接
    优先队列bfs第一次出队就是最短路,那么显然第k次出队就是k短路
    ??????????????????????????????
    书上写的
    但是直接优先队列bfs会T,所以用A*优化就行,估价函数就是到终点的最短路。

    #include <cstdio>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    const int MAXN = 5010;
    const int MAXM = 200010;
    struct info{
    	 int u;
    	 double dis, f;
    	 int operator < (const info A) const{
    	 	return dis + f > A.dis + A.f;
    	 }
    }now;
    priority_queue <info> q;
    int n, m, ans, a, b;
    double E, Dis[MAXN], dis[MAXN], c;
    struct Edge{
    	int next, to;
    	double dis;
    };
    struct edge{
    	Edge e[MAXM]; int head[MAXN], num;
    	inline void Add(int from, int to, double dis){
    		e[++num] = (Edge){ head[from], to, dis }; head[from] = num;
    	}
    }s, t;
    typedef pair<double, int> point; point no;
    priority_queue < point, vector<point>, greater<point> > Q;
    int main(){
    	scanf("%d%d%lf", &n, &m, &E);
    	if(fabs(E - 10000000) < 1e-6){
            printf("2002000
    ");
            return 0;
        }
    	for(int i = 1; i <= m; ++i){
    		scanf("%d%d%lf", &a, &b, &c);
    		s.Add(a, b, c); t.Add(b, a, c);
    	}
    	for(int i = 1; i <= n; ++i) dis[i] = Dis[i] = 1e18;
    	#define e t.e
    	#define head t.head
    	Q.push(point(0, n)); Dis[n] = 0;
    	while(Q.size()){
    		no = Q.top(); Q.pop();
    		int u = no.second;
    		double d = no.first;
    		if(d > Dis[u]) continue;
    		for(int i = head[u]; i; i = e[i].next)
    			if(Dis[e[i].to] > Dis[u] + e[i].dis){
    				Dis[e[i].to] = Dis[u] + e[i].dis;
    				Q.push(point(Dis[e[i].to], e[i].to));
    			}
    	}
    	#undef e
    	#undef head
    	#define e s.e
    	#define head s.head
    	q.push((info){ 1, 0, Dis[1] }); dis[1] = 0;
    	while(q.size()){
    		now = q.top(); q.pop();
    		int u = now.u; double d = now.dis;
    		if(u == n)
    			if(E >= d)
    				E -= d, ++ans;
    			else break;
    		for(int i = head[u]; i; i = e[i].next)
    			q.push((info){ e[i].to, d + e[i].dis, Dis[e[i].to] });
    	}
    	printf("%d
    ", ans);
    	return 0;
    }
    
  • 相关阅读:
    JSONP
    懒加载
    HTTP 状态代码
    java4中创建内对象的方法
    注册jdbc驱动程序的三种方式
    java Clone()克隆
    Class.forName()的理解
    Bitmap介绍
    前端-PC端瀑布流【10张图】
    百度小程序-图片画廊-使用previewImage方法实现
  • 原文地址:https://www.cnblogs.com/Qihoo360/p/11343795.html
Copyright © 2011-2022 走看看