hdu3665-Seaside(SPFA,dijkstra,floyd)

Seaside

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1369    Accepted Submission(s): 984

Problem Description

XiaoY is living in a big city, there are N towns in it and some towns near the sea. All these towns are numbered from 0 to N-1 and XiaoY lives in the town numbered ’0’. There are some directed roads connecting them. It is guaranteed that you can reach any town
from the town numbered ’0’, but not all towns connect to each other by roads directly, and there is no ring in this city. One day, XiaoY want to go to the seaside, he asks you to help him find out the shortest way.

 

Input

There are several test cases. In each cases the first line contains an integer N (0<=N<=10), indicating the number of the towns. Then followed N blocks of data, in block-i there are two integers, Mi (0<=Mi<=N-1) and Pi, then Mi lines followed. Mi means there
are Mi roads beginning with the i-th town. Pi indicates whether the i-th town is near to the sea, Pi=0 means No, Pi=1 means Yes. In next Mi lines, each line contains two integers S_Mi and L_Mi, which means that the distance between the
i-th town and the S_Mi town is L_Mi.

 

Output

Each case takes one line, print the shortest length that XiaoY reach seaside.

 

Sample Input

5
1 0
1 1
2 0
2 3
3 1
1 1
4 100
0 1
0 1

 

Sample Output

2

 

最短路径。话不多说，SPFA，dijkstra，floyd各贡献一枚！

dijkstra：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int map[15][15],dis[1000],vis[1000],target[15];
int N;
void dijkstra()
{
	memset(dis,INF,sizeof(dis));
	int i,j;
	for(i=0;i<N;i++)
	{
		dis[i]=map[0][i];
		vis[i]=0;
	}
	dis[0]=0;
	vis[0]=1;
	for(i=0;i<N;i++)
	{
		int mi=INF,mark=-1;
		for(j=0;j<N;j++)
		{
			if(!vis[j]&&dis[j]<mi)
			{
				mi=dis[j];
				mark=j;
			}
		}
		if(mark==-1)
		break;
		vis[mark]=1;
		for(j=0;j<N;j++)
		{
			if(!vis[j]&&dis[j]>dis[mark]+map[mark][j])
			dis[j]=dis[mark]+map[mark][j];
		}
	}
}
int main()
{
	int n,a,b,d,i,j,m;
	while(scanf("%d",&N)!=EOF)
	{
		int t=0;
		memset(map,INF,sizeof(map));
		for(i=0;i<N;i++)
		{
			scanf("%d%d",&m,&n);
			if(n)
			target[t++]=i;
			for(j=0;j<m;j++)
			{
				scanf("%d%d",&a,&b);
				map[i][a]=map[a][i]=b;
			}
		}
		dijkstra();
		int s[15];
		for(i=0;i<t;i++)
			s[i]=dis[target[i]];
		sort(s,s+t);
		printf("%d\n",s[0]);
	}
	return 0;
}

floyd：

#include<cstdio>
#include<cstring>
#define INF 0x3f3f3f3f
using namespace std;
int target[1000],map[15][15];
int N;
void floyd()
{
	int i,k,j;
	for(k=0;k<N;k++)
	{
		for(i=0;i<N;i++)
		{
			if(map[k][i]==INF)
			continue;
			for(j=0;j<N;j++)
			{
				if(map[i][j]>map[i][k]+map[k][j])
				map[i][j]=map[i][k]+map[k][j];
			}
		}
	}
}
int main()
{
	int n,i,j,a,d,m;
	while(scanf("%d",&N)!=EOF)
	{
		int t=0;
		memset(map,INF,sizeof(map));
		for(i=0;i<N;i++)
		{
			scanf("%d%d",&m,&n);
			if(n)
			target[t++]=i;
			for(j=0;j<m;j++)
			{
				scanf("%d%d",&a,&d);
				map[i][a]=map[a][i]=d;
			}
		}
		floyd();
		int q=INF;
		for(i=0;i<t;i++)
			q=q>map[0][target[i]]?map[0][target[i]]:q;
		printf("%d\n",q);
	}
	return 0;
 } 

SPFA：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std;
int head[20],target[15],vis[20],dis[20];
int N,t,w;
struct node
{
	int from,to,weight,next;
}s[1000];
void spfa()
{
	memset(vis,0,sizeof(vis));
	memset(dis,INF,sizeof(dis));
	queue<int> q;
	q.push(0);
	vis[0]=1;
	dis[0]=0;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		dis[0]=0;
		for(int k=head[u];k!=-1;k=s[k].next)
		{
			int v=s[k].to;
			if(dis[v]>dis[u]+s[k].weight)
			{
				dis[v]=dis[u]+s[k].weight;
				if(!vis[v])
				{
					q.push(v);
					vis[v]=1;
				 }
			}
		}

	}
	int qw[1000];
	for(int i=0;i<t;i++)
	qw[i]=dis[target[i]];
	sort(qw,qw+t);
	printf("%d\n",qw[0]);
}
int main()
{
	int n,a,b,d,i,j,m;
	while(scanf("%d",&N)!=EOF)
	{
		 t=0,w=0;
		memset(head,-1,sizeof(head));
		for(i=0;i<N;i++)
		{
			scanf("%d%d",&m,&n);
			if(n)
			target[t++]=i;
			for(j=0;j<m;j++)
			{
				scanf("%d%d",&a,&b);
				s[w].from=i;
				s[w].to=a;
				s[w].weight=b;
				s[w].next=head[i];
				head[i]=w++;
				s[w].from=a;
				s[w].to=i;
				s[w].weight=b;
				s[w].next=head[a];
				head[a]=w++;
			}
		}
		spfa();
	}
	return 0;
}

三种方法。各有利弊，可是floyd无疑是最简单的，三重for循环即可了。只是时间复杂度挺高的，easy超时。但这个题，0ms没问题！

有什么问题希望大家指出。谢谢！