HDU 1874 畅通工程续（最短路/spfa Dijkstra 邻接矩阵+邻接表）

题目链接: 传送门

畅通工程续

Time Limit: 1000MS     Memory Limit: 65536K

Description

某省自从实行了很多年的畅通工程计划后，终于修建了很多路。不过路多了也不好，每次要从一个城镇到另一个城镇时，都有许多种道路方案可以选择，而某些方案要比另一些方案行走的距离要短很多。这让行人很困扰。
现在，已知起点和终点，请你计算出要从起点到终点，最短需要行走多少距离。

Input

本题目包含多组数据，请处理到文件结束。
每组数据第一行包含两个正整数N和M (0<N<200,0<M<1000)，分别代表现有城镇的数目和已修建的道路的数目。城镇分别以0～N-1编号。
接下来是M行道路信息。每一行有三个整数A,B,X(0<=A,B<N,A!=B,0<X<10000),表示城镇A和城镇B之间有一条长度为X的双向道路。
再接下一行有两个整数S,T(0<=S,T<N)，分别代表起点和终点。

Output

对于每组数据，请在一行里输出最短需要行走的距离。如果不存在从S到T的路线，就输出-1.

Sample Input

3 3
0 1 1
0 2 3
1 2 1
0 2
3 1
0 1 1
1 2

Sample Output

2
-1

思路

spfa+邻接表

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

struct Edge{
	int u,v,w,next;
}; 

Edge edge[2005];

const int INF = 0x3f3f3f3f;
int dis[2005];
int head[2005];
int vis[2005];

void spfa(int s)
{
	memset(dis,0x3f3f3f3f,sizeof(dis));
	memset(vis,0,sizeof(vis));
	queue<int>que;
	dis[s] = 0;
	vis[s] = 1;
	que.push(s);
	while (!que.empty())
	{
		int curval = que.front();
		que.pop();
		vis[curval] = 0;
		for (int i = head[curval];i != -1;i = edge[i].next)
		{
			if (dis[curval]+edge[i].w < dis[edge[i].v])
			{
				dis[edge[i].v] = dis[curval] + edge[i].w;
				if (!vis[edge[i].v])
				{
					vis[edge[i].v] = 1;
					que.push(edge[i].v);
				}
			}
		}
	}
}


int main()
{
	int N,M;
	while (~scanf("%d%d",&N,&M))
	{
		int u,v,w,s,t;
		memset(head,-1,sizeof(head));
		for (int i = 0;i < 2*M;i+=2)
		{
			scanf("%d%d%d",&u,&v,&w);
			edge[i].u = u;edge[i].v = v;edge[i].w = w;edge[i].next = head[u];head[u] = i;
			edge[i+1].u = v;edge[i+1].v = u;edge[i+1].w = w;edge[i+1].next = head[v];head[v] = i+1;
		}
		scanf("%d%d",&s,&t);
		spfa(s);
		printf("%d\n",dis[t] == INF?-1:dis[t]);
	}
	return 0;
}

spfa+邻接矩阵

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

const int INF = 0x3f3f3f3f;
const int MAX = 1005;
int edge[MAX][MAX];

void spfa(int s,int n,int t)
{
	int dis[MAX];
	bool vis[MAX];
	memset(dis,0x3f,sizeof(dis));
	memset(vis,false,sizeof(vis));
	queue<int>que;
	dis[s] = 0;
	que.push(s);
	vis[s] = true;
	while (!que.empty())
	{
		int curval = que.front();
		que.pop();
		vis[curval] = false;
		for (int i = 0;i < n;i++)
		{
			if (dis[curval] < dis[i] - edge[curval][i])
			{
				dis[i] = dis[curval] + edge[curval][i];
				if (!vis[i])
				{
					que.push(i);
					vis[i] = true;
				}
			}
		}
	}
	printf("%d\n",dis[t] == INF?-1:dis[t]);
}


int main()
{
	int N,M;
	while (~scanf("%d%d",&N,&M))
	{
		int u,v,w,S,T;
		for (int i = 0;i < N;i++)
		{
			for (int j = 0;j < i;j++)
			{
				if (i == j) edge[i][j] = 0;
				else edge[i][j] = edge[j][i] = INF;
			}
		}
		for (int i =0;i < M;i++)
		{
			scanf("%d%d%d",&u,&v,&w);
			edge[u][v] = edge[v][u] = min(w,edge[u][v]);
		}
		scanf("%d%d",&S,&T);
		spfa(S,N,T);
	}
	return 0;
} 

Dijkstra+邻接矩阵

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int MAX = 205;
const int INF = 0x3f3f3f3f;
int edge[MAX][MAX];

void Dijkstra(int s,int t,int n)
{
	bool vis[MAX];
	int dis[MAX];
	int min,pos;
	memset(vis,false,sizeof(vis));
	for (int i = 0;i < n;i++)
	{
		dis[i] = edge[s][i];
	}
	for (int i = 1;i < n;i++)
	{
		min = INF;
		for (int j = 0;j < n;j++)
		{
			if (dis[j] < min && !vis[j])
			{
				pos = j;
				min = dis[j];
			}
		}
		vis[pos] = true;
		for (int j = 0;j < n;j++)
		{
			if (dis[pos] + edge[pos][j] < dis[j])
			{
				dis[j] = dis[pos] + edge[pos][j];
			}
		}
	}
	printf("%d\n",dis[t] == INF?-1:dis[t]);
}


int main()
{
	int N,M,S,T;
	while (~scanf("%d%d",&N,&M))
	{
		int u,v,w;
		for (int i = 0;i < N;i++)
		{
			for (int j = 0;j < i;j++)
			{
				if (i == j)edge[i][j] = edge[j][i] = 0;
				else edge[i][j] = edge[j][i] = INF;
			}
		}
		for (int i = 0;i < M;i++)
		{
			scanf("%d%d%d",&u,&v,&w);
			if (w < edge[u][v])
			{
				edge[u][v] = edge[v][u] = w;
			}
		}
		scanf("%d%d",&S,&T);
		Dijkstra(S,T,N);
	}
	return 0;
}