• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>
            隨筆 - 87  文章 - 279  trackbacks - 0
            <2025年7月>
            293012345
            6789101112
            13141516171819
            20212223242526
            272829303112
            3456789

            潛心看書研究!

            常用鏈接

            留言簿(19)

            隨筆分類(81)

            文章分類(89)

            相冊

            ACM OJ

            My friends

            搜索

            •  

            積分與排名

            • 積分 - 218072
            • 排名 - 117

            最新評論

            閱讀排行榜

            評論排行榜

            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(1290) 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
              
            亚洲精品白浆高清久久久久久| 新狼窝色AV性久久久久久| 国产精品日韩深夜福利久久 | 久久久精品久久久久久| 欧美大战日韩91综合一区婷婷久久青草 | 久久综合色区| 久久国产热精品波多野结衣AV| 99久久99久久精品国产片| 久久婷婷午色综合夜啪| 久久久综合九色合综国产| 久久婷婷午色综合夜啪| 国产精品久久久久久久久鸭| 中文字幕亚洲综合久久菠萝蜜| 久久Av无码精品人妻系列| 亚洲欧美国产精品专区久久| 久久精品嫩草影院| 中文字幕乱码久久午夜| 国产综合精品久久亚洲| 精品精品国产自在久久高清| 亚洲av日韩精品久久久久久a| 精品乱码久久久久久夜夜嗨| 国产精品久久网| 亚洲精品高清国产一线久久| 亚洲精品WWW久久久久久| 久久九九久精品国产| 国产精品久久久久无码av| 无码国内精品久久人妻| 少妇无套内谢久久久久| 欧美久久亚洲精品| 亚洲精品国产第一综合99久久| 香蕉久久夜色精品国产小说| 久久青草国产手机看片福利盒子| 国产综合久久久久| 国产精品一区二区久久国产| 色偷偷久久一区二区三区| 久久婷婷五月综合色奶水99啪| 亚洲国产天堂久久综合| 久久久黄色大片| 国内精品久久久久久久久电影网| 亚洲а∨天堂久久精品9966| 欧美成人免费观看久久|