hdu 4122（RMQ）2011福州现场赛B题

题意：月饼店卖月饼有一些预定月饼店在2000年一月一号0点开张k小时。每小时做出来的月饼价格不一样，月饼能保存若干小时每小时花费都告诉你。让你求月饼店完成所有订单的最小成本。

思路：这道题本身思路不难，我们知道月饼保存费用和每天做月饼的费用，我们就能算出来每天做月饼的成本，然后在这个数组上RMQ就可以了。就是日起处理比较麻烦，具体还是见代码吧。

代码如下：

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>

using namespace std;

typedef pair<int, int> pii;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int LEN = 100010;
ll que[2*LEN], ad[2*LEN], tb[LEN];
int day[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
struct CS {
  char mon[5];
  int y, d, h, num;
};
CS con[5000+10];


bool jy(int y) {
  return ((y%4==0 && y%100!=0) || y%400==0);
}

int changedata(char mon[], int d, int y, int h) {
  int ret = 0;
  if(y < 2000) return INF;
  for(int i=2000; i<y; i++) {
    if(jy(i)) ret += 366;
    else ret += 365;
  }
  int mtag = -1000;
  if(!strcmp(mon, "Jan")) {
    mtag = 1;
  } else if(!strcmp(mon, "Feb")) {
    mtag = 2;
  } else if(!strcmp(mon, "Mar")) {
    mtag = 3;
  } else if(!strcmp(mon, "Apr")) {
    mtag = 4;
  } else if(!strcmp(mon, "May")) {
    mtag = 5;
  } else if(!strcmp(mon, "Jun")) {
    mtag = 6;
  } else if(!strcmp(mon, "Jul")) {
    mtag = 7;
  } else if(!strcmp(mon, "Aug")) {
    mtag = 8;
  } else if(!strcmp(mon, "Sep")) {
    mtag = 9;
  } else if(!strcmp(mon, "Oct")) {
    mtag = 10;
  } else if(!strcmp(mon, "Nov")) {
    mtag = 11;
  } else if(!strcmp(mon, "Dec")){
    mtag = 12;
  }
  for(int i=1; i<mtag; i++) {
    ret += day[i];
    if(i==2 && jy(y)) ret ++;
  }
  ret += d-1;
  ret *= 24;
  ret += h;
  return ret;
}

//RMQ
ll dp[LEN][20], mm[LEN];
void initRMQ(int n, ll b[]) {
  mm[0] = -1;
  for(int i=1; i<=n; i++) {
    mm[i] = ((i&(i-1)) == 0) ? mm[i-1]+1 : mm[i-1];
    dp[i][0] = b[i];
  }
  for(int j=1; j<=mm[n]; j++)
    for(int i=1; i+(1<<j)-1 <= n; i++)
      dp[i][j] = min(dp[i][j-1], dp[i+(1 << (j-1))][j-1]);
}

ll rmq(int x, int y) {
  ll k = mm[y-x+1];
  return min(dp[x][k], dp[y-(1 << k)+1][k]);
}

int main()
{
#ifdef LOCALL
  freopen("in.txt","r",stdin);
  //freopen("out","w",stdout);
#endif

  int save, cost, n, m;
  while(scanf("%d%d", &n, &m)!=EOF) {
    if(!n && !m) break;
    for(int i=0; i<n; i++) {
      scanf("%s%d%d%d%d", con[i].mon, &con[i].d, &con[i].y, &con[i].h, &con[i].num);
    }
    scanf("%d%d", &save, &cost);
    for(int i=0; i<m; i++) {
      scanf("%I64d", &que[i]);
      ad[i] = (m-i)*cost;
    }
    for(int i=0; i<m; i++) {
      tb[i+1] =  ad[i]+que[i];
    }
    initRMQ(m, tb);
    ll ans = 0;
    for(int i=0; i<n; i++) {
      int locd = changedata(con[i].mon, con[i].d, con[i].y, con[i].h);
      if(locd > m) {continue;}
      ans += (rmq((locd-save+1>=1?locd-save+1:1), (locd+1<=m?locd+1:m))-ad[locd])*con[i].num;
    }
    printf("%I64d
", ans);
  }
  return 0;
}