hdu 3631(Floyd变形)

Shortest Path

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team’s coach, Prof. GUO asked them to solve the following shortest-path problem.
There is a weighted directed multigraph G. And there are following two operations for the weighted directed multigraph:
(1) Mark a vertex in the graph.
(2) Find the shortest-path between two vertices only through marked vertices.
For it was the first time that LMY faced such a problem, she was very nervous. At this moment, YY decided to help LMY to analyze the shortest-path problem. With the help of YY, LMY solved the problem at once, admiring YY very much. Since then, when LMY meets problems, she always calls YY to analyze the problems for her. Of course, YY is very glad to help LMY. Finally, it is known to us all, YY and LMY become programming lovers.
Could you also solve the shortest-path problem?

 

Input

The input consists of multiple test cases. For each test case, the first line contains three integers N, M and Q, where N is the number of vertices in the given graph, N≤300; M is the number of arcs, M≤100000; and Q is the number of operations, Q ≤100000. All vertices are number as 0, 1, 2, … , N - 1, respectively. Initially all vertices are unmarked. Each of the next M lines describes an arc by three integers (x, y, c): initial vertex (x), terminal vertex (y), and the weight of the arc (c). (c > 0) Then each of the next Q lines describes an operation, where operation “0 x” represents that vertex x is marked, and operation “1 x y” finds the length of shortest-path between x and y only through marked vertices. There is a blank line between two consecutive test cases.
End of input is indicated by a line containing N = M = Q = 0.

 

Output

Start each test case with "Case #:" on a single line, where # is the case number starting from 1.
For operation “0 x”, if vertex x has been marked, output “ERROR! At point x”.
For operation “1 x y”, if vertex x or vertex y isn’t marked, output “ERROR! At path x to y”; if y isn’t reachable from x through marked vertices, output “No such path”; otherwise output the length of the shortest-path. The format is showed as sample output.
There is a blank line between two consecutive test cases.

 

Sample Input

5 10 10
1 2 6335
0 4 5725
3 3 6963
4 0 8146
1 2 9962
1 0 1943
2 1 2392
4 2 154
2 2 7422
1 3 9896
0 1
0 3
0 2
0 4
0 4
0 1
1 3 3
1 1 1
0 3
0 4
0 0 0

 

Sample Output

Case 1:
ERROR! At point 4
ERROR! At point 1
0
0
ERROR! At point 3
ERROR! At point 4


解题思路：Floyd算法，必须要深刻理解Floyd算法中最外层循环的作用，这里的话把最外层的循环拆开了，每次mark的点就以它为中转点做松弛操作。这里最令人纠结的地方就是里面两层循环是不需要判断是否被mark的，因为如果没有被mark，那么后面在去寻找最短路时是不会用到这个点的，简单的画个图就会清楚了。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 305;
const int inf = 0x3f3f3f3f;
int n,m,q,map[maxn][maxn];
bool mark[maxn];

void update(int u)
{
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < n; j++)
		{
			if(map[i][j] > map[i][u] + map[u][j])
				map[i][j] = map[i][u] + map[u][j];
		} 
	}
}

int main()
{
	int u,v,c,op,cas = 1;
	while(scanf("%d%d%d",&n,&m,&q),n+m+q)
	{
		memset(mark,false,sizeof(mark));
		memset(map,inf,sizeof(map));
		for(int i = 1; i <= m; i++)
		{
			scanf("%d%d%d",&u,&v,&c);
			map[u][v] = min(map[u][v],c);
		}
		for(int i = 0; i < n; i++) map[i][i] = 0;
		if(cas!=1)  
            printf("\n");
		printf("Case %d:\n",cas++);
		while(q--)
		{
			scanf("%d",&op);
			if(op == 0)
			{
				scanf("%d",&u);
				if(mark[u] == true)
					printf("ERROR! At point %d\n",u);
				else 
				{
					mark[u] = true;
					update(u);
				}
			}
			else 
			{
				scanf("%d%d",&u,&v);
				if(mark[u] == false || mark[v] == false)
					printf("ERROR! At path %d to %d\n",u,v);
				else if(map[u][v] == inf)
					printf("No such path\n");
				else printf("%d\n",map[u][v]);
			}
		}
	}
	return 0;
}