[POJ 1821] Fence

[题目链接]

        http://poj.org/problem?id=1821

[算法]

       单调队列优化DP

[代码]

        

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXM 110
#define MAXN 16010

int i,j,k,n,m;
int f[MAXM][MAXN];
deque< int > q;

struct info
{
    int l,p,s;    
} a[MAXM];

inline bool cmp(info a,info b)
{
    return a.s < b.s;
}

int main()
{
    
    scanf("%d%d",&n,&m);
    for (i = 1; i <= m; i++) scanf("%d%d%d",&a[i].l,&a[i].p,&a[i].s);
    sort(a+1,a+m+1,cmp);
    for (i = 1; i <= m; i++)
    {
        q.clear();
        for (k = max(a[i].s - a[i].l,0); k < a[i].s; k++)
        {
            while (!q.empty() && f[i-1][k] - a[i].p * k >= f[i-1][q.back()] - a[i].p * q.back()) q.pop_back();
            q.push_back(k);    
        }    
        for (j = 1; j <= n; j++)
        {
            f[i][j] = max(f[i-1][j],f[i][j-1]);
            if (j >= a[i].s)
            {
                while (!q.empty() && q.front() < j - a[i].l) q.pop_front();
                if (!q.empty()) f[i][j] = max(f[i][j],f[i-1][q.front()] - a[i].p * q.front() + a[i].p * j);
            }
        }
    }
    printf("%d\n",f[m][n]);
    
    return 0;
}