P2299Mzc和体委的争夺战题解

这是一道Dijkstra经典题目（裸题）

P2299Mzc和体委的争夺战

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代码思路：Dijkstra+链式前向星+优先队列+结构体

其实很简单的

重点：

if(vis[n])
    return ;

意思是找到了立即返回，这样可以节约许些时间

 

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代码核心：

inline void dijkstra(int u){
	priority_queue<node>q;
	memset(dist, INF,sizeof(dist));
	dist[u] = 0;
	q.push(node(u,dist[u]));
	while(!q.empty()){
		int v = q.top().u;
		q.pop();
		if(vis[n])
			return ;		
		if(vis[v])
			continue;
		vis[v] = 1;
       //这里我写复杂了
		for(int i = head[v];i;i = E[i].next){
			if(!vis[E[i].to]&&dist[E[i].to]>dist[v]+E[i].w){
				dist[E[i].to] = dist[v] + E[i].w;
				q.push(node(E[i].to,dist[E[i].to]));
			}
		}
	}
}

将Dijkstra稍微改一改，符合题意即可

 

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这是实现优先队列内部从小到大以dist为标准排序
运用了重载运算符

	bool operator < (const node &a)const {
		return dist > a.dist;
	}

 

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下面是完整代码

 

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int maxn = 2505;
const int INF = 0x3f3f3f3f;
inline int read(){
	int x = 0;char c = getchar();
	while(c<'0'||c > '9') c = getchar();
	while(c>='0'&& c <='9') x = (x<<1)+(x<<3)+(c^48),c = getchar();
	return x;
}
struct node{
	int u, dist;
	node(){};
	node(int u_,int dist_){
		u = u_,dist = dist_;
	}
	bool operator < (const node &a)const {
		return dist > a.dist;
	}
};
struct edge{
	int to,next,w;
}E[maxn*maxn];
int head[maxn];
int cnt;
inline void add(int u,int v,int w){
	E[++cnt].to = v;
	E[cnt].w = w;
	E[cnt].next = head[u];
	head[u] = cnt;
}
int n,m;
int dist[maxn];
bool vis[maxn];
inline void dijkstra(int u){
	priority_queue<node>q;
	memset(dist, INF,sizeof(dist));
	dist[u] = 0;
	q.push(node(u,dist[u]));
	while(!q.empty()){
		int v = q.top().u;
		q.pop();
		if(vis[n])
			return ;		
		if(vis[v])
			continue;
		vis[v] = 1;
       //这里我写复杂了
		for(int i = head[v];i;i = E[i].next){
			if(!vis[E[i].to]&&dist[E[i].to]>dist[v]+E[i].w){
				dist[E[i].to] = dist[v] + E[i].w;
				q.push(node(E[i].to,dist[E[i].to]));
			}
		}
	}
}
int main(){
	n = read(),m = read();
	int u,v,w;
	memset(head,-1,sizeof(head));
	for(int i = 0;i<m;i++){
		u = read(), v = read(), w = read();
		add(u,v,w);
		add(v,u,w);
	}
	dijkstra(1);
	cout << dist[n];
	return 0;
}

 

 

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下面是提交结果：

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