遞推DP UVA 607 Scheduling Lectures

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題目傳送門

題意：教授給學生上課，有n個主題，每個主題有ti時間，上課有兩個限制：1. 每個主題只能在一節課內講完，不能分開在多節課；2. 必須按主題順序講，不能打亂。一節課L時間，如果提前下課了，按照時間多少，學生會有不滿意度。問最少要幾節課講完主題，如果多種方案輸出不滿意度最小的

分析：dp[i]表示前i個主題最少要多少節課講完，那麼這個主題可能和上個主題在同一節課講或者多開新的一節課講，狀態轉移方程：看代碼；優先滿足節數少的情況

收穫：普通的遞推DP

 

代碼：

/************************************************* Author        :Running_Time* Created Time  :2015-8-29 15:48:09* File Name     :UVA_607.cpp ************************************************/#include <cstdio>#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cmath>#include <string>#include <vector>#include <queue>#include <deque>#include <stack>#include <list>#include <map>#include <set>#include <bitset>#include <cstdlib>#include <ctime>using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1typedef long long ll;const int N = 1e3 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;int dp[N], d[N], a[N];int n, L, C;int get_DI(int x) {    if (!x) return 0;    if (1 <= x && x <= 10)  return -C;    else    return (x - 10) * (x - 10);}int main(void)    {    int cas = 0;    while (scanf ("%d", &n) == 1)   {        if (!n) break;        scanf ("%d%d", &L, &C);        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);        dp[0] = d[0] = a[0] = 0;        for (int i=1; i<=n; ++i)    {            dp[i] = dp[i-1] + 1;    d[i] = d[i-1] + get_DI (L - a[i]);            int cur = 0;            for (int j=i; j>=1; --j)    {                cur += a[j];                if (cur > L)    break;                if (dp[j-1] + 1 < dp[i] || (dp[j-1] + 1 == dp[i] && d[j-1] + get_DI (L - cur) < d[i]))    {                    dp[i] = dp[j-1] + 1;  d[i] = d[j-1] + get_DI (L - cur);                }            }        }        if (cas)    puts ("");        printf ("Case %d:\n", ++cas);        printf ("Minimum number of lectures: %d\n", dp[n]);        printf ("Total dissatisfaction index: %d\n", d[n]);    }    return 0;}

　　

遞推DP UVA 607 Scheduling Lectures