PAT (Top Level) Practise 1003 Universal Travel Sites (35)

1003. Universal Travel Sites (35)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

After finishing her tour around the Earth, CYLL is now planning a universal travel sites development project. After a careful investigation, she has a list of capacities of all the satellite transportation stations in hand. To estimate a budget, she must know the minimum capacity that a planet station must have to guarantee that every space vessel can dock and download its passengers on arrival.

Input Specification:

Each input file contains one test case. For each case, the first line contains the names of the source and the destination planets, and a positive integer N (<=500). Then N lines follow, each in the format:

source_i destination_i capacity_i

where source_i and destination_i are the names of the satellites and the two involved planets, and capacity_i > 0 is the maximum number of passengers that can be transported at one pass from source_i to destination_i. Each name is a string of 3 uppercase characters chosen from {A-Z}, e.g., ZJU.

Note that the satellite transportation stations have no accommodation facilities for the passengers. Therefore none of the passengers can stay. Such a station will not allow arrivals of space vessels that contain more than its own capacity. It is guaranteed that the list contains neither the routes to the source planet nor that from the destination planet.

Output Specification:

For each test case, just print in one line the minimum capacity that a planet station must have to guarantee that every space vessel can dock and download its passengers on arrival.

Sample Input:

EAR MAR 11EAR AAA 300EAR BBB 400AAA BBB 100AAA CCC 400AAA MAR 300BBB DDD 400AAA DDD 400DDD AAA 100CCC MAR 400DDD CCC 200DDD MAR 300

Sample Output:

700

这个题意有点难懂，不过其实就是个最裸的网络最大流，效率低的ek算法也能过，这里我用的是dinic

#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>#include<map>#include<iostream>#include<string>#include<stack>#include<algorithm>#include<bitset>#include<functional>using namespace std;typedef unsigned long long ull;typedef long long LL;const int maxn = 1e5 + 10;const int INF = 0x7FFFFFFF;map<string, int> M;char x[maxn], y[maxn];int n, z, tot;struct MaxFlow{  const static int maxe = 1e6 + 10;    //边数  const static int maxp = 1e5 + 10;   //点数  const static int INF = 0x7FFFFFFF;  struct Edges  {    int x, f;    Edges(){}    Edges(int x, int f) :x(x), f(f){}  }edge[maxe];  int first[maxp], next[maxe], dis[maxp], tot, work[maxp], n;  void clear(int x){ n = x; tot = 0; for (int i = 0; i <= n; i++) first[i] = -1; }  void AddEdge(int s, int t, int f)  {    edge[tot] = Edges(t, 0); next[tot] = first[s]; first[s] = tot++;    edge[tot] = Edges(s, f); next[tot] = first[t]; first[t] = tot++;  }  bool bfs(int s, int t)  {    for (int i = 0; i <= n; i++) dis[i] = -1;    queue<int> p;    p.push(s);    dis[s] = 0;    while (!p.empty())    {      int q = p.front();    p.pop();      for (int i = first[q]; i != -1; i = next[i])      {        if (edge[i ^ 1].f&&dis[edge[i].x] == -1)        {          p.push(edge[i].x);          dis[edge[i].x] = dis[q] + 1;          if (dis[t] != -1) return true;        }      }    }    return false;  }  int dfs(int s, int t, int low)  {    if (s == t) return low;    for (int &i = work[s], x; i >= 0; i = next[i])    {      if (dis[s] + 1 == dis[edge[i].x] && edge[i ^ 1].f && (x = dfs(edge[i].x, t, min(low, edge[i ^ 1].f))))      {        edge[i].f += x;    edge[i ^ 1].f -= x;  return x;      }    }    return 0;  }  int dinic(int s, int t)  {    int maxflow = 0, inc = 0;    while (bfs(s, t))    {      for (int i = 0; i <= n; i++) work[i] = first[i];      while (inc = dfs(s, t, INF)) maxflow += inc;    }    return maxflow;  }}solve;int main(){  while (scanf("%s%s%d", x, y, &n) != EOF)  {    solve.clear(2 * n + 2);    M[x] = 1;  M[y] = 2;  tot = 3;    while (n--)    {      scanf("%s%s%d", x, y, &z);      if (!M[x]) M[x] = tot++;      if (!M[y]) M[y] = tot++;      solve.AddEdge(M[x], M[y], z);    }    printf("%d\n", solve.dinic(1, 2));  }  return 0;}