[USACO07NOV] Milking Time

题目链接

动态规划转化成 DAG 然后拓扑求解的思路

虽然很简单不过感觉这个新思路会很有用！

如果两个事件互不影响并且有先后关系，就可以连一条有向边，跑最长路可以得到最后的最优解

实际上这还是个背包……

#include <queue>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 1000 + 10;
int n, m, r, head[maxn], dis[maxn], in_deg[maxn], edge_num, ans;

struct Segment {
  int lef, rig, enr;
  inline bool operator < (const Segment &x) const{ return this->rig < x.rig; }
} seg[maxn];

struct Edge { int v, nxt; } edge[maxn * maxn];

inline int read() {
  register char ch = 0; register int w = 0, x = 0;
  while( !isdigit(ch) ) w |= (ch == '-'),ch = getchar();
  while( isdigit(ch) ) x = (x * 10) + (ch ^ 48), ch = getchar();
  return w ? -x : x;
}

inline void Add_edge(int u, int v) {
  edge[++edge_num].v = v;
  edge[edge_num].nxt  = head[u], head[u] = edge_num;
}

inline void Breath_fs() {
  queue<int> q;
  for(int i = 1; i <= m; ++i) dis[i] = seg[i].enr;
  for(int i = 1; i <= m; ++i) if( !in_deg[i] ) q.push(i);
  while( !q.empty() ) {
    int x = q.front(); q.pop();
    for(int i = head[x]; i; i = edge[i].nxt) {
      if( dis[edge[i].v] < dis[x] + seg[edge[i].v].enr )
        dis[edge[i].v] = dis[x] + seg[edge[i].v].enr;
      if( --in_deg[edge[i].v] == 0 ) q.push(edge[i].v);
    }
  }
}

int main(int argc, char const *argv[])
{
  freopen("nanjolno.in", "r", stdin);
  freopen("nanjolno.out", "w", stdout);

  scanf("%d%d%d", &n, &m, &r);
  for(int i = 1; i <= m; ++i)
    seg[i].lef = read(), seg[i].rig = read() + r, seg[i].enr = read();
  sort(seg + 1, seg + m + 1), ans = 0;
  for(int i = 1; i <= m; ++i) for(int j = i + 1; j <= m; ++j)
      if( seg[i].rig <= seg[j].lef ) Add_edge(i, j), ++in_deg[j];
  Breath_fs();
  for(int i = 1; i <= m; ++i) ans = max(ans, dis[i]);
  printf("%d
", ans);

  fclose(stdin), fclose(stdout);
  return 0;
}

—— 冬天来了 —— 那是悲喜交加，永远的瞬间。