POJ 3159 Candies (Dijstra optimized non-vector notation)

Title Link: http://poj.org/problem?id=3159

Test instructions: give n personal candy, give the M group of data, each group of data contains a,b,c three number, meaning a candy number than B less than the number of C, that is, B candy number-a candy number <= C.

The last to ask N than 1 more than the number of sweets.

Can be from the condition of the

B-a<=c and B<=a+c finally to achieve this condition is to be b>a+c when the b=a+c can

So it's almost the shortest way. There are some optimizations in this question, such as

if (Vis[u]) continue; This avoids duplicate lookups of U-Points

#include <iostream> #include <string> #include <cstring> #include <vector> #include <queue > #include <cstdio> #define INF 0x3f3f3f3fusing namespace std;const int m = 2e5;int N, m, A, B, C, dis[30010]; struct TnT {int u, V, W, next;}    t[m];struct qnode{int V, c;    Qnode (int v, int c): V (v), C (c) {} BOOL operator < (const qnode &AMP;R) const{return c > r.c;    }};int head[30010], e;void Add (int u, int v, int w) {t[e].v = v;    T[E].W = W;    T[e].next = Head[u]; Head[u] = e++;}    BOOL Vis[m];void dij (int s) {priority_queue<qnode>q;    Memset (Vis, false, sizeof (VIS));    Q.push (Qnode (s, 0));    Dis[s] = 0;        while (!q.empty ()) {int u = q.top (). V;        Q.pop ();        if (Vis[u]) continue;        Vis[u] = true;            for (int i = head[u]; i =-1; i = t[i].next) {int v = t[i].v, w = T[I].W; if (!vis[v] && dis[v] > Dis[u] + W) {Dis[v] = dIs[u] + W;            Q.push (Qnode (V, Dis[v]));        }}}}int Main () {while (scanf ("%d%d", &n, &m)! = EOF) {e = 0;            for (int i = 1; I <= n; i++) {dis[i] = inf;        Head[i] =-1;            } for (int i = 1; I <= m; i++) {scanf ("%d%d%d", &a, &b, &c);        Add (A, B, c);        } dij (1);    printf ("%d\n", Dis[n]); } return 0;}

Poj 3159 Candies (Dijstra optimized non-vector notation)