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            寫了個比較通用的堆,可直接用作優(yōu)先隊列

            Silver Cow Party
            Time Limit:2000MS  Memory Limit:65536K
            Total Submit:1112 Accepted:326

            Description

            One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ XN). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.

            Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

            Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

             

            Input
            Line 1: Three space-separated integers, respectively: N, M, and X
            Lines 2..M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.

            Output
            Line 1: One integer: the maximum of time any one cow must walk.

            Sample Input

            4 8 2
            1 2 4
            1 3 2
            1 4 7
            2 1 1
            2 3 5
            3 1 2
            3 4 4
            4 2 3

             

            Sample Output

            10

             

            Hint
            Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

            Source
            USACO 2007 February Silver



            #include <iostream>
            using namespace std;

            const int INF = 1 << 28;

            int adj[1001][1001], adjw[1001][1001], na[1001];
            int n, m, x;


            //heap sink,swim,getmin,insert參數(shù)均為外部編號,wt為其權(quán)值
            int heap[100001], id[100001], hsize;
            int *key;
            void init(int s, int *wt) {
                
            int i;
                hsize 
            = s; 
                key 
            = wt;
                
            for (i=1; i<=hsize; i++{
                    heap[i] 
            = i;
                    id[i] 
            = i;
                }

            }

            void swim(int u) {
                
            int p = id[u], q = p >> 1, ku = key[u];
                
            while (q && ku < key[heap[q]]) {
                    id[heap[q]] 
            = p;
                    heap[p] 
            = heap[q];
                    p 
            = q;
                    q 
            = p >> 1;
                }

                id[u] 
            = p;
                heap[p] 
            = u;
            }

            void sink(int u) {
                
            int p = id[u],q = p << 1, ku = key[u];
                
            while (q <= hsize) {
                    
            if (q < hsize && key[heap[q+1]] < key[heap[q]]) q++;
                    
            if (key[heap[q]] >= ku) break;
                    id[heap[q]] 
            = p;
                    heap[p] 
            = heap[q];
                    p 
            = q; 
                    q 
            = p << 1;
                }

                id[u] 
            = p;
                heap[p] 
            = u;
            }

            int getmin() {
                
            int ret = heap[1];
                id[ret] 
            = -1;
                id[heap[hsize]] 
            = 1;
                heap[
            1= heap[hsize];
                hsize
            --;
                sink(heap[
            1]);
                
            return ret;
            }

            void insert(int u) {
                heap[
            ++hsize] = u;
                id[u] 
            = hsize;
                swim(u);
            }

            void build() {
                
            int i;
                
            for (i=hsize/2; i>0; i--) sink(heap[i]);
            }

            bool isEmpty() {
                
            return hsize == 0;
            }

            int dijkstraHeap(int beg, int end=-1{
                
            int i, j, k, u, v, w;
                
            int dist[1001], chk[1001];
                
            for (i=1; i<=n; i++{
                    dist[i] 
            = INF;
                    chk[i] 
            = 0;
                }

                init(n, dist);
                dist[beg] 
            = 0; swim(beg);
                
            while (!isEmpty()) {
                    u 
            = getmin();
                    
            if (u == end) break;
                    chk[u] 
            = 1;
                    
            for (i=0; i<na[u]; i++{
                        v 
            = adj[u][i];
                        w 
            = adjw[u][i];
                        
            if (dist[v] > dist[u] + w) {
                            dist[v] 
            = dist[u] + w;
                            swim(v);
                        }

                    }

                }

                
            if (end == -1return dist[n];
                
            return dist[end];
            }


            int main() {
                
            int i, j, k, u, v, w;
                
            int val[1001];
                scanf(
            "%d%d%d"&n, &m, &x);
                
            for (i=0; i<m; i++{
                    scanf(
            "%d%d%d"&u, &v, &w);
                    adj[u][na[u]] 
            = v; 
                    adjw[u][na[u]] 
            = w;
                    na[u]
            ++;
                }

               
                dijkstraHeap(x);
                memcpy(val, key, 
            sizeof(val));
                
                
            int ans = 0;
                
            for (i=1; i<=n; i++{
                    
            int tmp = dijkstraHeap(i,x);
                    
            if (tmp+val[i] > ans) ans = tmp + val[i];
                }

                
                printf(
            "%d\n", ans);
                
            return 0;
            }
            posted on 2007-07-23 20:51 閱讀(1289) 評論(4)  編輯 收藏 引用 所屬分類: 數(shù)據(jù)結(jié)構(gòu)與算法ACM題目

            FeedBack:
            # re: pku3268 dij+heap 2007-07-27 08:41 oyjpart
            終于更新blog了。。。  回復(fù)  更多評論
              
            # re: pku3268 dij+heap 2007-08-01 20:29 relic
            不必n次dijkstra,只要把所有邊反向,再來一次dijkstra就可以了。算上第一次一共兩次dij  回復(fù)  更多評論
              
            # re: pku3268 dij+heap 2007-08-03 22:58 
            偷懶了:)  回復(fù)  更多評論
              
            # re: pku3268 dij+heap 2007-09-18 13:16 drizzlecrj
            @relic
            re  回復(fù)  更多評論
              
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