• Hdu5385 The path


    The path

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 995    Accepted Submission(s): 419
    Special Judge

    Problem Description
    You have a connected directed graph.Let d(x) be the length of the shortest path from 1 to x.Specially d(1)=0.A graph is good if there exist x satisfy d(1)<d(2)<....d(x)>d(x+1)>...d(n).Now you need to set the length of every edge satisfy that the graph is good.Specially,if d(1)<d(2)<..d(n),the graph is good too.
    The length of one edge must  [1,n]
    It's guaranteed that there exists solution.
    Input
    There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
    The first line contains two integers n and m,the number of vertexs and the number of edges.Next m lines contain two integers each, ui and vi (1ui,vin), indicating there is a link between nodes ui and vi and the direction is from ui to vi.
    n3105,m6105
    1n,m105
    Output
    For each test case,print m lines.The i-th line includes one integer:the length of edge from ui to vi
    Sample Input
    2 4 6 1 2 2 4 1 3 1 2 2 2 2 3 4 6 1 2 2 3 1 4 2 1 2 1 2 1
    Sample Output
    1 2 2 1 4 4 1 1 3 4 4 4
    Author
    SXYZ
    Source
    题目大意:给定一张n个点m条有向边的图,构造每条边的边权(边权为正整数),令d(x)表示1到x的最短路,使得存在点i(1<=i<=n)满足d(1)<d(2)<…<d(i)>d(i+1)>…>d(n)。
    分析:有关最短路的构造题,和hdu4903有点类似,都是事先确定最短路再来构造边权.
       首先,令每个点的最短路都在[0,n-1]区间内,并且每个点的最短路都不一样.如果能够能够使所有点的最短路都合理,那么对于一条边所连的点u,v,只需要这条边的权值等于abs(d(v) - d(u))即可.这就能满足最短路的要求.
       接下来是要使得最短路合理.从两边向中间构造,维护两个指针l,r和一个计数器cnt,一开始l = 2,r = n.先设d[1] = 0,将1号点连边指向的点都打上标记.接着从l,r指针指向的点中选一个打上标记的点,那么这个点的d值就是++cnt.然后把这个点连边所指向的点都打上标记,并且向中间移动指针.因为题目保证有解,所以每次必然有一个指针能够移动.
       这样做是使得这些点到1的最短路合理,如果直接胡乱分配d,设定边权,那么可能只是这条边是最短路上的边,但实际上这些点到1的最短路并不一定是分配好的d.
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <cmath>
    #include <algorithm>
    
    using namespace std;
    
    const int maxn = 600010;
    
    int T,n,m,head[maxn],nextt[maxn],to[maxn],tot = 1,d[maxn],cnt;
    bool vis[maxn];
    
    struct node
    {
        int x,y,w;
    }e[600010];
    
    void add(int x,int y)
    {
        to[tot] = y;
        nextt[tot] = head[x];
        head[x] = tot++;
    }
    
    int main()
    {
        scanf("%d",&T);
        while (T--)
        {
            memset(head,0,sizeof(head));
            memset(vis,false,sizeof(vis));
            memset(d,0,sizeof(d));
            cnt = 1;
            tot = 1;
            scanf("%d%d",&n,&m);
            for (int i = 1; i <= m; i++)
            {
                scanf("%d%d",&e[i].x,&e[i].y);
                add(e[i].x,e[i].y);
            }
            d[1] = 0;
            for (int i = head[1];i;i = nextt[i])
            {
                int v = to[i];
                vis[v] = 1;
            }
            int l = 2,r = n;
            while (cnt < n)
            {
                int u;
                if (vis[l])
                    u = l++;
                else
                    u = r--;
                d[u] = cnt++;
                for (int i = head[u];i;i = nextt[i])
                {
                    int v = to[i];
                    vis[v] = 1;
                }
            }
            for (int i = 1; i <= m; i++)
            {
                if(e[i].x == e[i].y)
                    printf("%d
    ",n);
                else
                    printf("%d
    ",abs(d[e[i].x] - d[e[i].y]));
            }
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zbtrs/p/8453375.html
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