[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 =  + ;
 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 = ; register int w = , x = ;
   while( !isdigit(ch) ) w |= (ch == '-'),ch = getchar();
   while( isdigit(ch) ) x = (x * ) + (ch ^ ), 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 = ; i <= m; ++i) dis[i] = seg[i].enr;
   for(int i = ; 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] ==  ) 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 = ; i <= m; ++i)
     seg[i].lef = read(), seg[i].rig = read() + r, seg[i].enr = read();
   sort(seg + , seg + m + ), ans = ;
   for(int i = ; i <= m; ++i) for(int j = i + ; j <= m; ++j)
       if( seg[i].rig <= seg[j].lef ) Add_edge(i, j), ++in_deg[j];
   Breath_fs();
   for(int i = ; i <= m; ++i) ans = max(ans, dis[i]);
   printf("%d\n", ans);

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

—— 冬天来了 —— 那是悲喜交加，永远的瞬间。