题意:n个人发糖果,B 比 A 多 C的糖果,问最后第n个人比第一个人多多少的糖果
分析:最短路,Dijkstra 优先队列优化可过,SPFA竟然要用栈,队列超时!
代码:
/************************************************ * Author :Running_Time * Created Time :2015-9-1 19:18:52 * File Name :POJ_3159.cpp ************************************************/ #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 3e4 + 10; const int E = 150000 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; struct Edge { int v, w, nex; Edge (int _v = 0, int _w = 0) : v (_v), w (_w) {} bool operator < (const Edge &r) const { return w > r.w; } }edge[E]; int d[N]; int head[N]; int vis[N]; int cnt[N]; int n, m, e; void init(void) { memset (head, -1, sizeof (head)); e = 0; } void Dijkstra(int s) { memset (vis, false, sizeof (vis)); for (int i=1; i<=n; ++i) d[i] = INF; d[s] = 0; priority_queue<Edge> Q; Q.push (Edge (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; i=edge[i].nex) { int v = edge[i].v, w = edge[i].w; if (!vis[v] && d[v] > d[u] + w) { d[v] = d[u] + w; Q.push (Edge (v, d[v])); } } } } void SPFA(int s) { memset (vis, false, sizeof (vis)); memset (d, INF, sizeof (d)); d[s] = 0; vis[s] = true; stack<int> S; S.push (s); while (!S.empty ()) { int u = S.top (); S.pop (); vis[u] = false; for (int i=head[u]; ~i; i=edge[i].nex) { int v = edge[i].v, w = edge[i].w; if (d[v] > d[u] + w) { d[v] = d[u] + w; if (!vis[v]) { vis[v] = true; S.push (v); } } } } } void add_edge(int u, int v, int w) { edge[e].v = v, edge[e].w = w; edge[e].nex = head[u]; head[u] = e++; } int main(void) { while (scanf ("%d%d", &n, &m) == 2) { init (); for (int u, v, w, i=1; i<=m; ++i) { scanf ("%d%d%d", &u, &v, &w); add_edge (u, v, w); } //Dijkstra (1); SPFA (1); printf ("%d\n", d[n]); } return 0; }
SPFA/Dijkstra POJ 3159 Candies
原文:http://www.cnblogs.com/Running-Time/p/4776758.html