题意是给A和B发糖果,B的糖果数 – A的糖果数 <= c, 也就是B <= A + c,最后求n比1最多多
几个糖果。题目只有这一个约束条件,建图不难。将AB看成有向图的边,然后c看成边的权值,转化成
最短路来求解,大牛们都说了SPFA + queue会超时,所以用了SPFA + stack。因为这道题没有负
权的边,也可以用堆优化的dij来求这个最短路。
SPFA + Stack
/*Accepted 2396K 532MS C++ 1363B 2012-08-06 15:32:00*/#include#include #include 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#include #include #include 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 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 , greater > 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;}