UVa 11167 Monkeys in the Emei Mountain (最大流)

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题意:雪雪是一只猴子。它在每天的 2:00 —— 9:00之间非常渴,所以在这个期间它必须喝掉2个单位的水。它可以多次喝水,只要它喝水的总量是2.它从不多喝,在一小时内他只能喝一个单位的水。所以它喝水的时间段可能是2:00 ——4:00,或者3:00——5:00,或者7:00——9:00.甚至喝两次,第一次2:00——3:00,第二次8:00——9:00.但是它不能在1:00——3:00喝水,因为在1:00时它不渴,也不能在8:00——10:00喝水,因为9:00必须结束。一共有n(n <= 100)只这样的猴子。我们用一个(v,a,b)(0 <= v,a,b <= 50000,a < b,v <= b - a)来描述一个在时间a~b之间口渴,并且必须在这个期间喝够v个单位水的猴子。所有猴子喝水的速度都是1小时喝1个单位的水。现在的问题是只有一个地方可以让m只猴子同时喝水,需要作出一个能满足所有猴子的喝水需要的安排,问能否给出安排。多解时输出任意一组解即可。

析:很明显的把每只猴子都看作一个结点,然后把每个时间也是看成一个结点,但是,,,太多了吧,肯定会爆炸的,所以,先把所有的区间进行离散化,然后对每个区间建立结点,在输出解的时候,要注意,把所有的区间进行合并,只要对每个区间都加一个标志,循环标记,每次都加一个段,因为最大流最大的情况就是把整个区间都充满,不会超过。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-3;
const int maxn = 500 + 10;
const int maxm = 1e7 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Edge{
  int from, to, cap, flow;
};
struct Dinic{
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  int d[maxn];
  bool vis[maxn];
  int cur[maxn];

  void init(int n){
    this-> n = n;
    for(int i = 0; i < n; ++i)  G[i].cl;
    edges.cl;
  }

  void addEdge(int from, int to, int cap){
    edges.pb((Edge){from, to, cap, 0});
    edges.pb((Edge){to, from, 0, 0});
    m = edges.sz;
    G[from].pb(m - 2);
    G[to].pb(m - 1);
  }

  bool bfs(){
    ms(vis, 0);  vis[s] = 1;  d[s] = 0;
    queue<int> q;  q.push(s);

    while(!q.empty()){
      int u = q.front();  q.pop();
      for(int i = 0; i < G[u].sz; ++i){
        Edge &e = edges[G[u][i]];
        if(!vis[e.to] && e.cap > e.flow){
          vis[e.to] = 1;
          d[e.to] = d[u] + 1;
          q.push(e.to);
        }
      }
    }
    return vis[t];
  }

  int dfs(int u, int a){
    if(u == t || a == 0)  return a;
    int flow = 0, f;
    for(int &i = cur[u]; i < G[u].sz; ++i){
      Edge &e = edges[G[u][i]];
      if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
        e.flow += f;
        edges[G[u][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(a == 0)  break;
      }
    }
    return flow;
  }

  int maxflow(int s, int t){
    this-> s = s;
    this-> t = t;
    int flow = 0;
    while(bfs()){ ms(cur, 0);  flow += dfs(s, INF); }
    return flow;
  }
};

Dinic dinic;
vector<int> val;
int v[maxn], a[maxn], b[maxn];
int cnt[maxn];

int getpos(int x){ return lower_bound(val.begin(), val.end(), x) - val.begin(); }

int main(){
  int kase = 0;
  while(scanf("%d", &n) == 1 && n){
    scanf("%d", &m);
    val.cl;  val.pb(-1);
    int sum = 0;
    for(int i = 1; i <= n; ++i){
      scanf("%d %d %d", v+i, a+i, b+i);
      val.pb(a[i]);  val.pb(b[i]);
      sum += v[i];
    }
    sort(val.begin(), val.end());
    val.resize(unique(val.begin(), val.end()) - val.begin());
    int len = val.sz;
    int s = 0, t = len + n + 3;
    dinic.init(t + 5);
    for(int i = 1; i <= n; ++i){
      dinic.addEdge(s, len+i, v[i]);
      int l = getpos(a[i]);
      int r = getpos(b[i]);
      for(int j = l; j < r; ++j)  dinic.addEdge(len+i, j, val[j+1]-val[j]);
    }
    for(int j = 1; j + 1 < len; ++j)  dinic.addEdge(j, t, (val[j+1]-val[j])*m);
    if(sum != dinic.maxflow(s, t))  printf("Case %d: No\n", ++kase);
    else{
      printf("Case %d: Yes\n", ++kase);
      for(int i = 0; i < len; ++i)  cnt[i] = val[i];
      for(int i = 1; i <= n; ++i){
        vector<P> ans;
        for(int j = 0; j < dinic.G[i+len].sz && v[i]; ++j){
          Edge &e = dinic.edges[dinic.G[i+len][j]];
          if(e.from != i + len)  continue;
          int x = dinic.edges[dinic.G[i+len][j]^1].flow;
          if(x >= 0)  continue;
          x = -x;  v[i] -= x;
          if(val[e.to] + x < val[e.to+1]){
            if(x + cnt[e.to] <= val[e.to+1]){
              ans.push_back(P(cnt[e.to], cnt[e.to] + x));
              cnt[e.to] += x;
              if(cnt[e.to] == val[e.to+1])  cnt[e.to] = val[e.to];
            }
            else{
              x -= val[e.to+1] - cnt[e.to];
              ans.push_back(P(cnt[e.to], val[e.to+1]));
              cnt[e.to] = val[e.to] + x;
              ans.push_back(P(val[e.to], cnt[e.to]));
            }
          }
          else  ans.push_back(P(val[e.to], val[e.to+1]));
        }
        sort(ans.begin(), ans.end());
        for(int j = 0; j + 1 < ans.sz; ){
          if(ans[j].se == ans[j+1].fi)  ans[j].se = ans[j+1].se, ans.erase(ans.begin()+j+1);
          else ++j;
        }
        printf("%d", ans.sz);
        for(int j = 0; j < ans.sz; ++j)  printf(" (%d,%d)", ans[j].fi, ans[j].se);
        printf("\n");
      }
    }
  }
  return 0;
}

  

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