poj 1847 dijkstra(搬道岔的最少次数)

题意：节点i能够通向若干节点，但其中只有一个节点是直接连通的，其余的需要搬道岔。给定起点和终点，为最少的搬道岔次数是多少？如果不能到达输出-1

思路：构图时，直接连通的权值设为0，需要搬道岔的权值设置为1，用dijkstra即可。

dijkstra可以用堆来实现，复杂度O（ElogV）。

#include <stdio.h>
#include <string.h>
#define N 102
#define INF N
struct edge{
	int y,w,next;
}e[N*N];
int a,b,n;
int first[N],dis[N],visited[N],top;
void init(){
	int i;
	top = 0;
	memset(visited,0,sizeof(visited));
	memset(first,-1,sizeof(first));
	for(i = 0;i<=n;i++)
		dis[i] = INF;
}
void add(int x,int y,int w){
	e[top].y = y;
	e[top].w = w;
	e[top].next = first[x];
	first[x] = top++;
}
int dijkstra(){
	int i,min,pos;
	visited[a] = 1;
	for(i = first[a];i!=-1;i=e[i].next)
		dis[e[i].y] = e[i].w;
	while(1){
		min = INF;
		for(i = 1;i<=n;i++)
			if(!visited[i] && dis[i]<min){
				min = dis[i];
				pos = i;
			}
		if(min == INF)//不再能找到最小值
			return -1;
		visited[pos] = 1;
		if(pos == b)
			return dis[b];
		for(i = first[pos];i!=-1;i=e[i].next)
			if(!visited[e[i].y] && dis[pos]+e[i].w<dis[e[i].y])
				dis[e[i].y] = dis[pos]+e[i].w;
	}
}

int main(){
	freopen("a.txt","r",stdin);
	while(scanf("%d %d %d",&n,&a,&b)!=EOF){
		int i,j,k,num;
		init();
		for(i = 1;i<=n;i++){
			scanf("%d",&num);
			for(j = 0;j<num;j++){
				scanf("%d",&k);
				if(!j)//直接相连
					add(i,k,0);
				else//需要搬道岔
					add(i,k,1);
			}
		}
		printf("%d\n",dijkstra());
	}
	return 0;
}

堆版本：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <cstdlib>
using namespace std;
#define INF 0x3fffffff
#define N 105
struct edge{
    int y,next,w;
    bool operator<(const struct edge &b)const{
        return w>b.w;
    }
}e[N*N*2];
int first[N],top,n,a,b,used[N],dis[N];
priority_queue<struct edge> h;
void add(int x,int y,int w){
    e[top].w = w;
    e[top].y = y;
    e[top].next = first[x];
    first[x] = top++;
}
int dijkstra(){
    int i,j;
    memset(used, 0, sizeof(used));
    for(i = 1;i<=n;i++)
        dis[i] = INF;
    dis[a] = 0;
    used[a] = 1;
    for(i = first[a];i!=-1;i=e[i].next){
        h.push(e[i]);
        dis[e[i].y] = e[i].w;
    }
    while(!h.empty()){
        struct edge now = h.top(),tmp;
        h.pop();
        if(used[now.y])
            continue;
        used[now.y] = 1;
        if(now.y == b)
            break;
        for(j = first[now.y];j!=-1;j=e[j].next){
            if(!used[e[j].y] && dis[e[j].y]>dis[now.y]+e[j].w){
                tmp.y = e[j].y;
                tmp.w = dis[now.y] + e[j].w;
                dis[e[j].y] = dis[now.y]+e[j].w;
                h.push(tmp);//一定注意此处：插入堆中的是更新后的从源点到e[j].y的最短路径长度
            }
        }
    }
    if(dis[b] == INF)
        return -1;
    return dis[b];
}
int main(){
    int i,j,num;
    memset(first,-1,sizeof(first));
    top = 0;
    scanf("%d %d %d",&n,&a,&b);
    for(i = 1;i<=n;i++){
        scanf("%d",&num);
        if(num){
            scanf("%d",&j);
            add(i,j,0);
            num--;
        }
        while(num--){
            scanf("%d",&j);
            add(i,j,1);
        }
    }
    printf("%d\n",dijkstra());
    return 0;
}