[USACO07NOV] Milking Time

Posted 2021-01-24 nanjoqin

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……

 1 #include <queue>
 2 #include <cstdio>
 3 #include <cctype>
 4 #include <cstring>
 5 #include <iostream>
 6 #include <algorithm>
 7 using namespace std;
 8 
 9 const int maxn = 1000 + 10;
10 int n, m, r, head[maxn], dis[maxn], in_deg[maxn], edge_num, ans;
11 
12 struct Segment {
13   int lef, rig, enr;
14   inline bool operator < (const Segment &x) const{ return this->rig < x.rig; }
15 } seg[maxn];
16 
17 struct Edge { int v, nxt; } edge[maxn * maxn];
18 
19 inline int read() {
20   register char ch = 0; register int w = 0, x = 0;
21   while( !isdigit(ch) ) w |= (ch == ‘-‘),ch = getchar();
22   while( isdigit(ch) ) x = (x * 10) + (ch ^ 48), ch = getchar();
23   return w ? -x : x;
24 }
25 
26 inline void Add_edge(int u, int v) {
27   edge[++edge_num].v = v;
28   edge[edge_num].nxt  = head[u], head[u] = edge_num;
29 }
30 
31 inline void Breath_fs() {
32   queue<int> q;
33   for(int i = 1; i <= m; ++i) dis[i] = seg[i].enr;
34   for(int i = 1; i <= m; ++i) if( !in_deg[i] ) q.push(i);
35   while( !q.empty() ) {
36     int x = q.front(); q.pop();
37     for(int i = head[x]; i; i = edge[i].nxt) {
38       if( dis[edge[i].v] < dis[x] + seg[edge[i].v].enr )
39         dis[edge[i].v] = dis[x] + seg[edge[i].v].enr;
40       if( --in_deg[edge[i].v] == 0 ) q.push(edge[i].v);
41     }
42   }
43 }
44 
45 int main(int argc, char const *argv[])
46 {
47   freopen("nanjolno.in", "r", stdin);
48   freopen("nanjolno.out", "w", stdout);
49 
50   scanf("%d%d%d", &n, &m, &r);
51   for(int i = 1; i <= m; ++i)
52     seg[i].lef = read(), seg[i].rig = read() + r, seg[i].enr = read();
53   sort(seg + 1, seg + m + 1), ans = 0;
54   for(int i = 1; i <= m; ++i) for(int j = i + 1; j <= m; ++j)
55       if( seg[i].rig <= seg[j].lef ) Add_edge(i, j), ++in_deg[j];
56   Breath_fs();
57   for(int i = 1; i <= m; ++i) ans = max(ans, dis[i]);
58   printf("%d
", ans);
59 
60   fclose(stdin), fclose(stdout);
61   return 0;
62 }

 

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