Layout(POJ 3169)

Posted 2021-01-02 ymir-taomee

• 原题如下：
  Layout

  Time Limit: 1000MS		Memory Limit: 65536K
  Total Submissions: 14484		Accepted: 6961

  Description

  Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and since they can be rather pushy, it is possible that two or more cows can line up at exactly the same location (that is, if we think of each cow as being located at some coordinate on a number line, then it is possible for two or more cows to share the same coordinate). 
  
  Some cows like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows dislike each other and the minimum distance by which they must be separated. 
  
  Your job is to compute, if possible, the maximum possible distance between cow 1 and cow N that satisfies the distance constraints.

  Input

  Line 1: Three space-separated integers: N, ML, and MD. 
  
  Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart. 
  
  Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.

  Output

  Line 1: A single integer. If no line-up is possible, output -1. If cows 1 and N can be arbitrarily far apart, output -2. Otherwise output the greatest possible distance between cows 1 and N.

  Sample Input

  4 2 1
  1 3 10
  2 4 20
  2 3 3

  Sample Output

  27

  Hint

  Explanation of the sample: 
  
  There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart. 
  
  The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.
• 题解：记第i号牛的位置d[i]，首先，牛是按照编号顺序排列的，所以有d[i]≤d[i+1]成立，其次，对于每对关系好的牛之间的最大距离限制，都有d[AL]+DL≥d[BL]成立，同样对于每对关系不好的牛，都有d[AD]+DD≤d[BD]成立，因此，原问题就转化为了在满足这三类不等式的情况下，求解d[N]-d[1]的最大值的问题。这种线性规划问题当然可以用单纯形法来求，但这道题还比较特殊，这些不等式的两边都只有1个变量，这种特殊形式的不等式方程组又叫做差分约束系统，它可以用我们熟悉的图论最短路算法来求。最短路问题可以用这种不等式的形式表示出来，记从起点s出发，到各个顶点v的最短距离为d(v)，因此，对于每条权值为w的边e=(v,u)，都有d(v)+w≥d(u)成立，在满足全部这些约束不等式中的d中，d(v)-d(s)的最大值就是从s到v的最短距离。对比差分约束与最短路问题，两者具有完全一样的形式，所以，可以把原来的问题的每一个约束不等式对应成图中的一条边来构图，然后通过解决最短路问题来解决原问题。首先把顶点编号为1~N，d[i]≤d[i+1]变形为d[i+1]+0≥d[i]，因此，从顶点i+1向顶点i连一条权值为0的边，同样d[AL]+DL≥d[BL]对应从顶点AL向顶点BL连一条权值为DL的边，d[AD]+DD≤d[BD]对应从BD向顶点AD连一条权值为-DD的边。所求的问题是d[N]-d[1]的最大值，对应为顶点1到顶点N的最短距离。
• 代码：

    1 #include <cstdio>
    2 #include <queue>
    3 #include <cctype>
    4 #define num s-‘0‘
    5 using namespace std;
    6 
    7 struct edge
    8 {
    9     int to;
   10     int cost;
   11 };
   12 
   13 const int INF=0x3f3f3f3f;
   14 const int MAX_V=1000;
   15 vector<edge> G[MAX_V];
   16 int V, ML, MD;
   17 int d[MAX_V];
   18 bool inque[MAX_V];
   19 int n[MAX_V];
   20 
   21 void read(int &x){
   22     char s;
   23     x=0;
   24     bool flag=0;
   25     while(!isdigit(s=getchar()))
   26         (s==‘-‘)&&(flag=true);
   27     for(x=num;isdigit(s=getchar());x=x*10+num);
   28     (flag)&&(x=-x);
   29 }
   30 
   31 void write(int x)
   32 {
   33     if(x<0)
   34     {
   35         putchar(‘-‘);
   36         x=-x;
   37     }
   38     if(x>9)
   39         write(x/10);
   40     putchar(x%10+‘0‘);
   41 }
   42 
   43 bool spfa();
   44 
   45 int main()
   46 {
   47     read(V); read(ML); read(MD);
   48     for (int i=0; i<ML; i++)
   49     {
   50         int A, B, C;
   51         read(A); read(B); read(C);
   52         edge e;
   53         e.to=B-1;e.cost=C;
   54         G[A-1].push_back(e);
   55     }
   56     for (int i=0; i<MD; i++)
   57     {
   58         int A, B, C;
   59         read(A); read(B); read(C);
   60         edge e;
   61         e.to=A-1;e.cost=-C;
   62         G[B-1].push_back(e);
   63     }
   64     for (int i=0; i<V-1; i++)
   65     {
   66         edge e;
   67         e.to=i;e.cost=0;
   68         G[i+1].push_back(e);
   69     }
   70     if (spfa())
   71     {
   72         if (d[V-1]==INF) write(-2);
   73         else write(d[V-1]);
   74     }
   75     else write(-1);
   76 }
   77 
   78 bool spfa()
   79 {
   80     fill(d, d+V, INF);
   81     fill(inque, inque+V, false);
   82     fill(n, n+V, 0);
   83     queue<int> que;
   84     d[0]=0;
   85     que.push(0);
   86     n[0]++;
   87     inque[0]=true;
   88     while (!que.empty())
   89     {
   90         int v=que.front();que.pop();
   91         inque[v]=false;
   92         for (int i=0; i<G[v].size(); i++)
   93         {
   94             if (d[G[v][i].to]>d[v]+G[v][i].cost)
   95             {
   96                 d[G[v][i].to]=d[v]+G[v][i].cost;
   97                 if (!inque[G[v][i].to]) 
   98                 {
   99                     que.push(G[v][i].to);
  100                     inque[G[v][i].to]=true;
  101                     if (++n[G[v][i].to]>V) return false;
  102                 }
  103             }
  104         }
  105     }
  106     return true;
  107 }