hdu 2851 Beat It 单调队列+DP

最新推荐文章于 2019-08-29 09:17:35 发布

kongming_acm

最新推荐文章于 2019-08-29 09:17:35 发布

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分类专栏： acm_数据结构 acm_动态规划文章标签： integer each output input ini

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本文介绍了一款在人人开放平台上的怪物攻击游戏背景及算法实现。游戏中，玩家需利用特殊大炮攻击怪物，考虑日夜伤害差异及大炮冷却时间等因素，通过算法计算出最大伤害值。

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Problem Description

There's a monster-attacking game on the Renren's open platform, and the game's background are showing below.
Where lies a strange country, lices a strange monster. The strange people in that strange country are persecuted by the strange monster. So teh strange people make a strange cannon to defeat the strange monster.
As far as we know, the strange cannon is powerful, but once it works, it should not work in next T hours (Eg. If it fired at time 1, then time 2 to time T it cannot work). When the strange monster get hit during the daylight, it get PD point hurt; it will get PN point hurt when it happens during the night. As the country is strange, the time of the daylight and night changes eveyday. In the ith day, the daylight starts at time 1 and ends at time T1[i], then from time (T1[i]+1) to time (T1[i]+T2[i]) is night.
Here comes a question: what is the maximal hurt point would the strange monster got after N days?

Input

First line contains an integer Q (1<=Q<=20), indicates the number of test cases.
For each case, first line contains 4 integers: N (1<=N<=1000), T (1<=T<=100), PD, PN.
Then followed N lines, the (i+1)th line contains 2 integer T1[i], T2[i].
(PD, PN, T1[i], T2[i] are all non-negative and fit the 32bit integer)

Output

For each case print one line, output the answer. See the sample for format.

Sample Input

Sample Output

Case 1: 65

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=700000;
__int64 f[maxn];//f[i]表示以i结尾且第i个小时打炮的最大伤害
//f[i]=max(f[k])+pd|pn;1<=k<=i-t;
bool d[maxn];
__int64 Q[maxn];
int I[maxn];
int main()
{
    int ci,pl=1;scanf("%d",&ci);
    while(ci--)
    {
        int n,t;
        __int64 pd,pn;
        scanf("%d%d%I64d%I64d",&n,&t,&pd,&pn);
        __int64 extra=0;
        int head=1,tail=0;
        int m=0;//小时
        for(int i=0;i<n;i++)
        {
            __int64 t1,t2;scanf("%I64d%I64d",&t1,&t2);
            __int64 tmp;
            if(t1>t+t) tmp=(t1-t-t)/t;
            else tmp=0;
            extra+=tmp*pd;
            for(int j=0;j<t1-tmp*t;j++)
            {
                d[++m]=1;
            }
            if(t2>=t+t) tmp=(t2-t-t)/t;
            else tmp=0;
            extra+=tmp*pn;
            for(int j=0;j<t2-tmp*t;j++)
            {
                d[++m]=0;
            }
        }
        __int64 res=0;
        for(int i=1;i<=m;i++)
        {
            if(i-t>=1) //f[i-t] 入队
            {
                while(head<=tail&&Q[tail]<=f[i-t]) tail--;//a[i]入队
                tail++;
                Q[tail]=f[i-t],I[tail]=i-t;
                f[i]=Q[head]+(d[i]?pd:pn);
            }
            else f[i]=(d[i]?pd:pn);
            res=max(res,f[i]);
        }
        printf("Case %d: %I64d\n",pl++,extra+res);
    }
    return 0;
}