POJ3616 -- Milking Time

POJ3616 -- Milking Time
Milking Time

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:4964		Accepted:2076

Description

Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her nextN(1 ≤N≤0. 1,000,000) hours (conveniently labeled 0 ..N-1) so that she produces as much milk as possible.

Farmer John has a listM(1 ≤MLess than or equal to 1,000) possibly overlapping intervals in which he is available for milking. Each intervalIHas a starting hour (0 ≤Starting_houri≤N), An ending hour (Starting_houri<Ending_houri≤N), And a corresponding efficiency (1 ≤EfficiencyiLess than or equal to 1,000,000) which indicates how many gallons of milk that he can get out of Bessie in that interval. farmer John starts and stops milking at the beginning of the starting hour and ending hour, respectively. when being milked, Bessie must be milked through an entire interval.

Even Bessie has her limitations, though. After being milked during any interval, she must restR(1 ≤R≤N) Hours before she can start milking again. Given Farmer Johns list of intervals, determine the maximum amount of milk that Bessie can produce inNHours.

Input

* Line 1: Three space-separated integers:N,M, AndR
* Lines 2 ..M+ 1: LineI+ 1 describes FJ's ith milking interval withthree space-separated integers:Starting_houri,Ending_houri, AndEfficiencyi

Output

* Line 1: The maximum number of gallons of milk that Bessie can product inNHours

Sample Input

12 4 21 2 810 12 193 6 247 10 31

Sample Output

43

Source

USACO 2007 November Silver

Dp, uses the memory-based search, and creates an edge based on the meaning of the question.

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            #include using namespace std;const int N = 1000010;int dp[1010];int head[1010];int w[1010];int tot;struct kode{int next;int to;}edge[1010 * 1010];struct node{int l, r, eff;}hour[1010];void addedge(int from, int to){edge[tot].to = to;edge[tot].next = head[from];head[from] = tot++;}int dfs(int u){if (dp[u]){return dp[u];}for (int i = head[u]; ~i; i = edge[i].next){int v = edge[i].to;dp[u] = max(dp[u], dfs(v));}dp[u] += w[u];return dp[u];}int main(){int n, m, r;while (~scanf("%d%d%d", &n, &m, &r)){memset (head, -1, sizeof(head));memset (dp, 0, sizeof(dp));tot = 0;for (int i = 1; i <= m; ++i){scanf("%d%d%d", &hour[i].l, &hour[i].r, &hour[i].eff);w[i] = hour[i].eff;}for (int i = 1; i <= m; ++i){for (int j = i + 1; j <= m; ++j){if (hour[i].r + r <= hour[j].l){addedge(i, j);// printf("%d -- %d\n", i, j);}else if (hour[j].r + r <= hour[i].l){addedge(j, i);// printf("%d -- %d\n", j, i);}}}int ans = 0;for (int i = 1; i <= m; ++i){ans = max(ans, dfs(i));}printf("%d\n", ans);}return 0;}