• <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>
            posts - 7,comments - 3,trackbacks - 0
            Electricity

            Time Limit: 10 Seconds      Memory Limit: 32768 KB      Special Judge

            After recent blackouts in some regions in North America, the government has decided to reorganize the power supply network of the continent.

            The power supply network is the set of nodes, such as power plants or transformation stations, connected by transmission lines. All lines are used to transmit electricity from one node to another. For stability reasons the system is organized in such a way that there are no directed cycles.

            Since the government is currently short of money due to several small peaceful militaristic operations, it cannot build new power lines for the moment. So after reorganization the same lines will be used, but some lines will have to transmit electricity in the direction opposite to the current one. To make the reorganization gentle enough, the management of the power network is planning to switch the transmission direction for exactly one line each day. Of course, no day there must be a cycle in a network, since this may cause damage to the system. The resulting network is also designed to be acyclic.

            Help them to plan the reorganization.

            Input

            There are mutilple cases in the input file.

            The first line of the input file contains n --- the number of nodes in the network, and m --- the number of transmission lines (2 <= n <= 1,000 , 1 <= m <= 10,000 ). The following m lines contain three integer numbers each. The first two give the source and the destination node for the corresponding line in the current node. The third number is 0 if the line must keep its transmission direction in the resulting network, and 1 if the direction must be reversed.

            There can be several lines connecting the same pair of nodes, but due to acyclicity condition, they all transmit electricity in the same direction. This is also the reason why no line connects a node to itself.

            There is an empty line after each case.

            Output

            First output k --- the number of days in the plan you suggest. You don't need to minimize this number, but it must not exceed 4m . After that print k integer numbers --- for each day output the number of the line that changes the transmission direction this day. If it is impossible to make the desired reorganization, output -1 instead of k .

            There should be an empty line after each case.

            Sample Input

            4 5
            1 2 0
            2 3 1
            2 4 1
            1 4 1
            4 3 0
            
            2 2
            1 2 1
            1 2 1
            
            

            Sample Output

            3
            3 2 4
            
            -1
            
            

            Source: Andrew Stankevich's Contest #9


            一道不錯的圖論題,大意是給一個有向網絡N,必須將其中X條邊反向,給出一個方案順序,使得在執行反向的過程中,網絡不會出現環,如果不存在這個方案,輸出-1.

            思路:首先,判斷不存在的情況:1.兩個重邊都要反轉,那么在操作過程中肯定出現環;2.在初始網絡或結束網絡存在環。對于第二種情況,可以用topo排序求一下兩個網絡是否存在環,并記錄下topp序列。
            之后,我們要構造方案,構造方案的過程有兩個:
            1.一個點的可逆入度邊(這句是廢話,可逆邊就是必須改變的邊,肯定包含于方案,但注意是入度)
            2.這個點必須從結尾網絡的topo序列從后往前搜索.....(凌亂了吧....)
            首先,我們要明白topo序列的性質,就是序列a1,a2,a3,...,an,表示的是網絡n個點的線性關系,存在任意的i<j,使得ai -> aj,也就是,如果用網絡表示topo序列,那么只有往右邊指向的邊。
            通過這一個性質,加上網絡N前后兩次的topo序列,我們不難發現,結尾網絡topo序列的最后一個肯定是在操作中失去出度而從初始的topo序列降為(或停留)最后面,所以,將其可反轉的入度邊反向,肯定不會存在回射邊從而產生環結構,因此,從最后一個點向前搜索,每次執行可執行的反轉操作,那么一定能保持當前點的不存在回射邊。
            對于一個點,可存在同時支配多條可反轉邊,因為答案要求一次一次執行反轉,如果對于同一個點順序不當可能出現環,所以我們考慮以下問題:x->y, x->z,如果反轉(x,y)從而導致了環的出現,那么可以肯定z->y是成立的,而在topo關系上z比y靠前,所以我們對于同一個點輸出結果時,要按照其初始網絡的topo順序,從左向右輸出。
            思路完畢,AC,證明....略了吧,我證明了一草稿紙。
            注意不要用矩陣,我因為那個吃了幾次CE......
            代碼:
            #include <iostream>
            #include 
            <cstdio>
            #include 
            <cstring>
            #include 
            <queue>
            #include 
            <vector>
            #define N 1100
            #define M 10010
            using namespace std;

            int n, m;

            struct edge
            {
                
            int u, v, next;
            } et[
            2][M];

            int eh[2][N], tot[2];
            int be[2 * N], ed[2 * N], sta[2 * N];
            int deg[2][N], deg2[N];
            int g[N][N];

            void add(int u, int v, int i)
            {
                
            int t = ++tot[i];
                et[i][t].u 
            = u;
                et[i][t].v 
            = v;
                et[i][t].next 
            = eh[i][u];
                eh[i][u] 
            = t;
                deg[i][v]
            ++;
            }

            int topo(int eh[], edge et[], int que[], int deg[])
            {
                
            int i, j, k, top = -1, qt = 0;
                
            for (i = 1; i <= n; ++i)
                    deg2[i] 
            = deg[i];
                
            for (i = 1; i <= n; ++i)
                    
            if (deg2[i] == 0)
                        sta[
            ++top] = i;
                
            for (j = 1; j <= n; ++j)
                {
                    
            if (top == -1return 0;
                    
            int u = sta[top--];
                    que[qt
            ++= u;
                    
            for (i = eh[u]; i != -1; i = et[i].next)
                    {
                        deg2[et[i].v]
            --;
                        
            if (deg2[et[i].v] == 0) sta[++top] = et[i].v;
                    }
                }
                
            return 1;
            }

            int was[M], cnt;
            void slove()
            {
                
            int i, j, k;
                memset(was, 
            0sizeof(was));
                printf(
            "%d\n", cnt);
                
            for (i = n - 1; i >= 0--i)
                {
                    
            int u = ed[i];
                    
            for (j = 0; j < n; ++j)
                      
            if (g[u][be[j]] > 0)
                      {
                          
            if (was[g[u][be[j]]] == 0)
                          {
                              was[g[u][be[j]]] 
            = 1;
                              printf(
            "%d ", g[u][be[j]]);
                          }
                      }
                }
                
            if (cnt > 0) printf("\n");
            }

            int main()
            {
                
            int i, j, k;
                
            while (scanf("%d%d"&n, &m) != EOF)
                {
                    memset(eh, 
            -1sizeof(eh));
                    memset(deg, 
            0sizeof(deg));
                    
            for (i = 1; i <= n; ++i)
                      
            for (j = 1; j <= n; ++j)
                        g[i][j] 
            = 0;
                    tot[
            1= tot[0= 0;
                    cnt 
            = 0;
                    
            int flag = 1;
                    
            for (i = 1; i <= m; ++i)
                    {
                        
            int a, b, c;
                        scanf(
            "%d%d%d"&a, &b, &c);
                        add(a, b, 
            0);
                        
            if (c)
                        {
                            add(b, a, 
            1);
                            cnt
            ++;
                            
            if (g[a][b] > 0) flag = 0;
                            g[a][b] 
            = i;
                        }
                        
            else
                          add(a, b, 
            1);
                    }
                    
            if (flag == 0)
                    {
                        printf(
            "-1\n\n");
                        
            continue;
                    }
                    
            int first = topo(eh[0], et[0], be, deg[0]);
                    
            int last = topo(eh[1], et[1], ed, deg[1]);
                    
            if (!(first && last))
                    {
                        printf(
            "-1\n\n");
                        
            continue;
                    }
                    slove();
                    printf(
            "\n");
                }
                
            return 0;
            }
            posted on 2011-10-15 22:12 LLawliet 閱讀(150) 評論(0)  編輯 收藏 引用 所屬分類: 圖論
            亚洲午夜久久久精品影院| 九九热久久免费视频| 狠狠色综合网站久久久久久久高清 | 久久国产视频网| 久久五月精品中文字幕| 日韩人妻无码精品久久免费一 | 色偷偷888欧美精品久久久| 久久99久久成人免费播放| 国产精品久久新婚兰兰| 亚洲精品乱码久久久久久自慰| 99精品国产在热久久无毒不卡| 国产精品久久久久久久久久免费| 亚洲欧美日韩精品久久亚洲区 | 亚洲国产精品久久久久久| 久久久午夜精品| 亚洲国产精品婷婷久久| 久久久久亚洲AV无码观看| 国产精品久久久天天影视香蕉 | 狠狠色伊人久久精品综合网| 狠狠综合久久AV一区二区三区| 精品国产婷婷久久久| 久久w5ww成w人免费| 99久久免费国产精品特黄| 久久91这里精品国产2020| 久久精品国产99国产电影网 | 久久精品中文无码资源站| 超级碰久久免费公开视频| 国产婷婷成人久久Av免费高清| 亚洲国产精品狼友中文久久久| 激情综合色综合久久综合| 精品久久一区二区三区| 久久精品国产亚洲AV香蕉| 欧美黑人激情性久久| 久久国产免费直播| 亚洲国产美女精品久久久久∴| 久久综合亚洲鲁鲁五月天| 久久久噜噜噜久久中文字幕色伊伊| 欧美粉嫩小泬久久久久久久| 久久久久国产精品嫩草影院| 久久久久婷婷| 久久青青草视频|