POJ 2387 Til the Cows Come Home （最短路径 模版题 三种解法）

原题链接：Til the Cows Come Home

题目大意：有  个点，给出从  点到  点的距离并且  和  是互相可以抵达的,问从  到  的最短距离。

题目分析：这是一道典型的最短路径模版题，需要注意的是：使用dijkstra算法求解需要考虑有重复边问题，而使用bellman-ford算法 和 spfa算法 可以忽略这个问题。

代码如下：

// Dijkstra
#include 
using namespace std;

const int INTFY = 1 << 29;
const int MAX   = 1005;
int map[MAX][MAX];
int n,m;
 
void dijkstra() {
	int min, v;
	int d[MAX];
	bool vis[MAX];
 
	for(int i = 1; i <= n; i++) {
		vis[i] = 0;
		d[i]   = map[1][i];
	}
 
	for(int i = 1; i <= n; i++) {
		min = INTFY;
		for(int j = 1; j <= n; j++)
			if(!vis[j] && d[j] < min) {
				v   = j;
				min = d[j];
			}
		vis[v] = 1;
 
		for(int j = 1; j <= n; j++)
			if(!vis[j] && d[j] > map[v][j] + d[v])
				d[j] = map[v][j] + d[v];
	}
	cout << d[n] << endl;
}
 
int main() {
	int a, b, c;
	while(cin >> m >> n) {
		for(int i = 1; i <= n; i++)
			for(int j = 1; j <= n; j++)
				if(i == j)
					map[i][i] = 0;
				else map[i][j] = map[j][i] = INTFY;
		
		for(int i = 1; i <= m; i++) {
			cin >> a >> b >> c;
			if(map[a][b] > c) map[a][b] = map[b][a] = c;
		}
		dijkstra();
	}
	return 0;
}

// Bellman-ford
#include 
using namespace std;

const int INFTY  = 1 << 29;
const int MAXM = 2005;
const int MAXV = 1005;
 
typedef struct {
	int a, b, w;
}Edge;
 
Edge edge[MAXM];
int n, m;
 
void bellman_ford() {
	int d[MAXV];
	for(int i = 2; i <= n; i++) d[i] = INFTY;
	d[1] = 0;
 
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= m; j++) {
			if(d[edge[j].a] > edge[j].w + d[edge[j].b]) d[edge[j].a] = edge[j].w + d[edge[j].b];
			if(d[edge[j].b] > edge[j].w + d[edge[j].a]) d[edge[j].b] = edge[j].w + d[edge[j].a];
		}
	}
	cout << d[n] << endl;
}
 
int main() {
	int a, b, c;
	while(cin >> m >> n) {
		for(int i = 1; i <= m; i++) {
			cin >> a >> b >> c;
			edge[i].a = a;
			edge[i].b = b;
			edge[i].w = c;
		}
		bellman_ford();
	}
	return 0;
}

// Spfa
#include 
#include 
using namespace std;

const int INFTY = 1 << 29;
const int MAXM = 4005;
const int MAXV = 1005;

typedef struct{
	int a, b, w, next;
}Edge;
 
Edge edge[MAXM];
int n, m, headlist[MAXV];
 
void spfa() {
	int d[MAXV], vis[MAXV];
	queue q;
	for(int i = 2; i <= n; i++) {
		d[i] = INFTY;
		vis[i] = 0;
	}
	
	d[1] = 0;
	vis[1] = 1;
	q.push(1);
	
	while(!q.empty()) {
		int v = q.front(); q.pop();
		vis[v] = 0;
 
		for(int i = headlist[v]; i != -1; i = edge[i].next)
			if(d[v] + edge[i].w < d[edge[i].b]) {
				d[edge[i].b] = d[v] + edge[i].w;
				if(!vis[edge[i].b]) {
					vis[edge[i].b] = 1;
					q.push(edge[i].b);
				}
			}
	}
	cout << d[n] << endl;
}
 
int main() {
	int a, b, c;
	while(cin >> m >> n) { 
		for(int i = 1; i <= n; i++) headlist[i] = -1;
		for(int i = 1; i <= 2 * m; i += 2) {
			cin >> a >> b >> c;
			edge[i].a = a;
			edge[i].b = b;
			edge[i].w = c;
			edge[i].next = headlist[a];
			headlist[a] = i;
			edge[i+1].a = b;
			edge[i+1].b = a;
			edge[i+1].w = c;
			edge[i+1].next = headlist[b];
			headlist[b] = i + 1;
		}
		spfa();
	}
	return 0;
}