POJ 1511 Invitation Cards (Dijstra optimization)

Title Link: http://poj.org/problem?id=1511

Test instructions: gives n points and n a forward edge, all points to the source point 1 back and forth the shortest possible sum (guaranteed each point can return to the source point 1)

The problem is relatively simple is the number of sides and points a bit more so you can use dijstra+ priority queue so complexity can be to V*logn

#include <iostream> #include <cstring> #include <string> #include <vector> #include <queue  > #include <cstdio> #define INF 1000000000using namespace Std;const int M = 1e6 + 10;int N, M, a[m], b[m], c[m] , dis[m];struct TnT {int V, w;};    struct CMP {bool operator () (int x, int y) {return dis[x] > Dis[y];    }};vector<tnt>vc[m];bool vis[m];void dij (int s) {priority_queue<int, vector<int>, cmp>q;    Memset (Vis, false, sizeof (VIS));    TnT GG;    Q.push (s);    Dis[s] = 0;        while (!q.empty ()) {int m = q.top ();        Vis[m] = true;            for (int i = 0; i < vc[m].size (); i++) {GG = vc[m][i];                if (Dis[m] + GG.W < DIS[GG.V]) {DIS[GG.V] = dis[m] + gg.w;                    if (!VIS[GG.V]) {VIS[GG.V] = true;                Q.push (GG.V);    }}} q.pop ();    }}int Main () {int t;    TnT GG; scanf ("%d", &AMP;T);        while (t--) {scanf ("%d%d", &n, &m);            for (int i = 1; I <= n; i++) {vc[i].clear ();        Dis[i] = inf;            } for (int i = 1; I <= m; i++) {scanf ("%d%d%d", &a[i], &b[i], &c[i]);            GG.V = B[i], GG.W = C[i];        Vc[a[i]].push_back (GG);        } dij (1);        A long long sum = 0;        for (int i = 1; I <= n; i++) {sum + = (long long) dis[i];            } for (int i = 1; I <= n; i++) {vc[i].clear ();        Dis[i] = inf;            } for (int i = 1; I <= m; i++) {gg.v = A[i], GG.W = C[i];        Vc[b[i]].push_back (GG);        } dij (1);        for (int i = 1; I <= n; i++) {sum + = (long long) dis[i];    } printf ("%lld\n", sum);    } return 0; }

Poj 1511 Invitation Cards (Dijstra optimization)