Sunday, July 31, 2011

UVa 11813 - Shopping

Problem Link:

Problem in short -
Given a weighted undirected graph with around 10^5 nodes and edges and at most 10 store nodes which you must visit, create the minimum distance route starting from a particular node that visits all other nodes and returns to that node.

To solve the problem, we'll need the shortest path between each pair of store nodes. To get that we'll run a dijkstra from each possible store node and store the distances to all other nodes. When we know distance between each pair of nodes, we can easily find the optimal route using a bitmask dp where state would be current node we're visiting and set of nodes we've already visited.

Source code -

#pragma warning(disable:4786)
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cstdlib>
#include <cstring>
using namespace std;

#define min(a,b) (((a)<(b))?(a):(b))
#define rep(i,n) for(i=0;i<(n);i++)
#define i64 long long
//#define i64 __int64
#define MAXN 100005
#define MAXS 11
typedef pair<int,int> ii;
typedef vector<ii> vii;

i64 INF;

int n,m;
int ns;
int s[MAXS];
vector< pair<int,int> > adj[MAXN];
priority_queue< ii, vii, greater<ii> > pq;
i64 d[MAXS][MAXN];

i64 dp[MAXS][(1<<MAXS)];

void dijkstra(int st, int dex) {
 int i;
 pair<int, int> t;
 rep(i, n) d[dex][i] = INF;
 d[dex][st] = 0;
 while(!pq.empty()) pq.pop();
 pq.push( make_pair(st, 0) );
 while(!pq.empty()) {
  t =;
  if(d[dex][t.first] >= t.second) {
   rep(i,adj[t.first].size()) {
    if(d[dex][t.first] + adj[t.first][i].second < d[dex][adj[t.first][i].first]) {
     d[dex][adj[t.first][i].first] = d[dex][t.first] + adj[t.first][i].second;
     pq.push( make_pair(adj[t.first][i].first, d[dex][adj[t.first][i].first]) );

i64 go(int cur, int state) {
 if(state == 0) return d[cur][ s[ns-1] ];
 if(dp[cur][state] != -1) return dp[cur][state];
 i64 &ret = dp[cur][state];
 int i;
 ret = INF;
 rep(i,ns) if( state & (1<<i) ) {
  ret = min(ret, go(i, state - (1<<i)) + d[cur][ s[i] ] );
 return ret;

int main() {
 int T;
 int u,v,d,i;
 scanf(" %d",&T);
 INF = (1<<30); INF*=INF;
 while(T--) {
  scanf(" %d %d",&n,&m);
  rep(i,n) adj[i].clear();
  rep(i,m) {
   scanf(" %d %d %d",&u, &v, &d);
   adj[u].push_back( make_pair(v,d) );
   adj[v].push_back( make_pair(u,d) );
  scanf(" %d",&ns);
  rep(i,ns) scanf(" %d",&s[i]);
  s[ns++] = 0;
  rep(i,ns) {
   dijkstra(s[i], i);

  memset(dp, -1, sizeof(dp));
  printf("%lld\n", go(ns-1, (1<<ns)-1) );
 return 0;


  1. hello,
    This is very wonderful blog it contain many fact and views about shopping...

  2. good one. But can you briefly explain the bit masking technique you have used there. i.e. go() method.

  3. Nice problem.........and nice solution

  4. Very Good Point about the your blog when I search in google your blog get ranking good all blog post are very informative. Kindly post about Interior Designs In Chennai & Modular Kitchen in Chennai related.