題意是給A和B發糖果，B的糖果數?–?A的糖果數?<=?c,?也就是B?<=?A?+?c，最后求n比1最多多

幾個糖果。題目只有這一個約束條件，建圖不難。將AB看成有向圖的邊，然后c看成邊的權值，轉化成

最短路來求解，大牛們都說了SPFA?+?queue會超時，所以用了SPFA?+?stack。因為這道題沒有負

權的邊，也可以用堆優化的dij來求這個最短路。

?

SPFA + Stack

/*Accepted 2396K 532MS C++ 1363B 2012-08-06 15:32:00*/ #include<cstdio> #include<cstring> #include<cstdlib>const int inf = 0x3f3f3f3f; const int V = 30005; const int E = 150005; int pnt[E], cost[E], nxt[E]; int head[V], e, dist[V]; bool vis[V];int relax( int u, int v, int c) {if( dist[v] > dist[u] + c) {dist[v] = dist[u] + c;return 1;}return 0; }void addedge( int u, int v, int c) {pnt[e] = v; cost[e] = c; nxt[e] = head[u]; head[u] = e ++; }int SPFA( int src, int n) {int i;for(i = 1; i <= n; ++ i){vis[i] = false; dist[i] = inf;}dist[src] = 0;int S[E], top = 1;S[0] = src; vis[src] = true;while(top) {int u, v;u = S[ --top]; vis[u] = false;for(i = head[u]; i != -1; i = nxt[i]) {v = pnt[i];if( 1 == relax( u, v, cost[i]) && !vis[v]) {S[ top ++] = v; vis[v] = true;}}}return dist[n]; }int main() {int n, m;while( scanf( "%d%d", &n, &m) == 2){int a, b, c;e = 0;memset( vis, false, sizeof vis);memset( head, -1, sizeof head);while( m --){scanf( "%d%d%d", &a, &b, &c);addedge( a, b, c);}printf( "%d\n", SPFA( 1, n));}return 0; }

?

?Dijkstra

/*Accepted 2392K 610MS C++ 1376B 2012-08-06 15:27:31*/ #include<stdio.h> #include<string.h> #include<stdlib.h> #include<queue> using namespace std;const int MAXN = 30030, MAXM = 150150; const int inf = 0x3f3f3f3f; int first[MAXN], next[MAXM], v[MAXM], w[MAXM], dist[MAXN]; int n, m, e; typedef pair<int, int> pii;void addedge(int a, int b, int c) {v[e] = b, w[e] = c;next[e] = first[a], first[a] = e ++; }void ReadGraph() {int a, b, c;e = 0;memset(first, -1, sizeof first);while(m --){scanf("%d%d%d", &a, &b, &c);addedge(a, b, c);} }int Dijkstra(int src, int n) {int i, x;pii u;priority_queue<pii, vector<pii>, greater<pii> > q;for(i = 1; i <= n; i ++) dist[i] = inf;dist[src] = 0;q.push(make_pair(dist[src], src));while(!q.empty()){u = q.top(), q.pop();x = u.second;if(dist[x] != u.first) continue;if(n == x) break;for(i = first[x]; i != -1; i = next[i]){if(dist[v[i]] > dist[x] + w[i]){dist[v[i]] = dist[x] + w[i];q.push(make_pair(dist[v[i]], v[i]));}}}return dist[n]; }int main() {while(scanf("%d%d", &n, &m) == 2){ReadGraph();printf("%d\n", Dijkstra(1, n));}return 0; }

?

?

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轉載于:https://www.cnblogs.com/Yu2012/archive/2012/08/06/2440719.html

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