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
Title effect: a given point is n, m have to FIG edges, the points are not added began to FIG. When asked then C times, enter 0 for joining the vertices if the vertices U has been added too, enter "ERROR! At point U", input 1, if one of these two points are not added, output "ERROR! At path U to V", if the distance between two points is infinite, the output "No such path", otherwise the output path of the two.
Ideas: as he said to simulate the like, as to how to add a little, if you understand the algorithm, then floyed should know that the outermost K is used to represent a short circuit can pass through most of the point, not the final, but also because the topic step by step adding points, so use floyed can be solved.
AC Code:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e3 + 5;
const int INF = 1e8;
int p[maxn][maxn], n, m, c, X;
bool vis[maxn], flag;
void init() {
for (int i = 0; i < n; i++) {
vis[i] = 0;
for (int j = 0; j < n; j++) {
if (i == j) p[i][j] = 0;
else p[i][j] = INF;
}
}
}
void floyed (int k) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
p[i][j] = min (p[i][j], p[i][k] + p[k][j]);
}
}
int main()
{
while (scanf("%d%d%d", &n, &m, &c) && n + m + c) {
if (flag) cout << endl;
flag = 1;
init();
for (int i = 0; i < m; i++) {
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
p[u][v] = min (p[u][v], w);
}
cout << "Case " << ++X << ":" << endl;
while (c--) {
int ai, bi, ci;
scanf("%d%d", &ai, &bi);
if (!ai) {
if (vis[bi]) cout << "ERROR! At point " << bi << endl;
else {
vis[bi] = 1;
floyed(bi);
}
}
else {
scanf("%d", &ci);
if (!vis[bi] || !vis[ci]) cout << "ERROR! At path " << bi << " to " << ci << endl;
else if (p[bi][ci] < INF) cout << p[bi][ci] << endl;
else cout << "No such path" << endl;
}
}
}
return 0;
}