- Floyd Warshall's এলগরিদম ব্যবহার করে খুব সহজেই এই সমস্যা সমাধান করা যায় ।
- আমাদের ইনপুট গ্রাফে যদি u থেকে v তে যাওয়ার জন্য একবার টাইম দেয়া থাকল ২০ সেকেন্ড পরে আবার u থেকে v তে যাওয়ার টাইম দেয়া থাকল ৩০ সেকেন্ড । সেই ক্ষেত্রে আমাদের কে মিনিমাম টাইম অর্থাৎ ২০ সেকেন্ড কে নিতে হবে ।
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
/** | |
* Problem : 341 - Non-Stop Travel. | |
* Verdict : Accepted. | |
* Writer : Mehadi Hasan Menon. | |
* Time : 0.00 ms. | |
**/ | |
#include <iostream> | |
#include <cstdio> | |
using namespace std; | |
const int inf = 100000007; | |
int region[12][12]; | |
int p[12][12]; | |
void floyd_warshall(int node) | |
{ | |
for(int k = 1; k <= node; k++) | |
{ | |
for(int i = 1; i <= node; i++) | |
{ | |
for(int j = 1; j <= node; j++) | |
{ | |
if(region[i][k] + region[k][j] < region[i][j]) | |
{ | |
region[i][j] = region[i][k] + region[k][j]; | |
p[i][j] = p[i][k]; | |
} | |
} | |
} | |
} | |
} | |
void print_path(int i, int j) | |
{ | |
printf(" %d", i); | |
while(i != j) | |
{ | |
i = p[i][j]; | |
printf(" %d", i); | |
} | |
} | |
int main() | |
{ | |
freopen("input.txt", "r+", stdin); | |
int ni, streets, u, v, time, tc; | |
tc = 1; | |
while(scanf("%d", &ni) && ni != 0) | |
{ | |
for(int i = 1; i <= ni; i++) | |
{ | |
for(int j = 1; j <= ni; j++) | |
{ | |
region[i][j] = inf; | |
} | |
} | |
for(u = 1; u <= ni; u++) | |
{ | |
scanf("%d", &streets); | |
for(int i = 0; i < streets; i++) | |
{ | |
scanf("%d %d", &v, &time); | |
// IMPORTANT :D | |
region[u][v] = min(region[u][v], time); | |
p[u][v] = v; | |
} | |
} | |
floyd_warshall(ni); | |
int src, des; | |
scanf("%d %d", &src, &des); | |
printf("Case %d: Path =", tc++); | |
print_path(src, des); | |
printf("; %d second delay\n", region[src][des]); | |
} | |
return 0; | |
} |
Comments
Post a Comment