1030. Travel Plan (30)

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (<=500) is the number of cities (and hence the cities are numbered from 0 to N-1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

City1 City2 Distance Cost

where the numbers are all integers no more than 500, and are separated by a space.

Output Specification:

For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

Sample Input

4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20

Sample Output

0 2 3 3 40

這一題和前面的有個很像，其關鍵思路就是用深度優先思路進行路線的查找然後回溯，直到找到最進的並且cost最低的那條路徑。爲了讓深搜的效率更高，我用了dijkstra算法先把最短的路徑求出來，這樣在每次遞歸調用dfs的時候優先進行深度比較，如果大於了最短路徑直接回溯。

代碼如下：

#include <stdio.h>

#define PRINT_I(x) printf("%s = %d\n",#x,x);
#define PRINT_F(x) printf("%s = %f\n",#x,x);
#define PRINT_S(x) printf("%s = %s\n",#x,x);

const int INFINITY = 10000000;

int path[500][500];
int cost[500][500];
int visit[500] = {0};
int dist[500];
int min_path,min_cost,min_count,count,fdist,fcost,border;
int result[500],tmp[500];

void init(int *v,int n);
int find_min(int *v,int n,int *visit);
void dijsktra(int s,int n);
int dfs(int s,int n,int d);

int main(int argc,char* argv[])
{
	int i,j,n,m,s,d;

	//freopen("input","r",stdin);
	scanf("%d%d%d%d",&n,&m,&s,&d);
	//initial path and cost
	for(i=0;i<n;i++)
		for(j=0;j<n;j++)
			path[i][j] = cost[i][j] = (i==j)?0:INFINITY;

	for(i=0;i<m;i++)
	{
		int x,y,distance,mcost;
		scanf("%d%d%d%d",&x,&y,&distance,&mcost);
		path[x][y] = path[y][x] = distance;
		cost[x][y] = cost[y][x] = mcost;	
	}

	dijsktra(s,n);
	border = dist[d];
	min_path = INFINITY;
	min_cost = INFINITY;

	count = fcost = fdist = 0;
	init(visit,n);
	visit[s] = 1;
	dfs(s,n,d);

	printf("%d ",s);
	for(i=0;i<min_count;i++)
		printf("%d ",result[i]);
	printf("%d %d\n",min_path,min_cost);

	return 0;
}

void init(int *v,int n)
{
	int i;
	for(i=0;i<n;i++)
		v[i] = 0;
}

int find_min(int *v,int n,int *visit)
{
	int i,min,min_index;

	min_index = -1;
	min = INFINITY;
	for(i=0;i<n;i++)
	{
		if(v[i]<min && !visit[i])
		{
			min = v[i];
			min_index = i;
		}
	}
	return min_index;
}

void dijsktra(int s,int n)
{
	int i,j,min;

	visit[s] = 1;
	for(i=0;i<n;i++)
		dist[i] = path[s][i];

	min = find_min(dist,n,visit);
	while(min != -1)
	{
		visit[min] = 1;
		for(j=0;j<n;j++)
		{
			if(path[s][min] + path[min][j] < path[s][j])
				dist[j] = path[s][min] + path[min][j];
		}
		min = find_min(dist,n,visit);
	}
}

int dfs(int s,int n,int d)
{
	int i;

	if(fdist > border)
		return 0;

	if(s == d)
	{
		if( (fdist<min_path) || (fdist==min_path&&fcost<min_cost) )
		{
			for(i=0;i<count;i++)
			 result[i] = tmp[i];
			min_count = count;
			min_path = fdist;
			min_cost = fcost;
		}
		else
			return 0;
	}

	for(i=0;i<n;i++)
	{
		if(path[s][i]!=INFINITY && !visit[i])
		{
			visit[i] = 1;
			fdist += path[s][i];
			fcost += cost[s][i];
			tmp[count++] = i;
			dfs(i,n,d);
			visit[i] = 0;
			fdist -= path[s][i];
			fcost -= cost[s][i];
			count--;
		}
	}

	return 0;
}