HDU 5653 Bomber Man wants to bomb an Array(dp)

题意:

$N\le 2\times 10^3,N个格子,M\le N个炸弹$
$每个炸弹可以向左向右炸任意距离,假设为L,R,那么贡献E_i=L+R+1$
$每个格子只能炸1次,总贡献为\Pi_{i=1}^mE_i$
$求最大的总贡献$

分析:

$dp,f[i]:=前i个格子被炸掉的最大贡献$
$转移就枚举炸弹,枚举左右炸的距离,然后这个看起来的三方的dp是二方的$
$f[0]=1,f[r] = max(f[r], f[l-1]*(r-l+1)),log2一下就变成+了$
$只看左右端点,都只是枚举了n$
$时间复杂度O(n^2)$

代码:

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//
// Created by TaoSama on 2016-04-07
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n, m;
int a[N];
double v[N], f[N];
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
for(int i = 1; i <= 2000; ++i) v[i] = log2(i);
while(t--) {
scanf("%d%d", &n, &m);
a[0] = 0; a[m + 1] = n + 1;
for(int i = 1; i <= m; ++i) scanf("%d", a + i), ++a[i];
sort(a + 1, a + 1 + m);
memset(f, 0, sizeof f);
f[0] = 0;
for(int i = 1; i <= m; ++i)
for(int l = a[i - 1] + 1; l <= a[i]; ++l)
for(int r = a[i]; r <= a[i + 1] - 1; ++r)
f[r] = max(f[r], f[l - 1] + v[r - l + 1]);
long long ans = 1e6 * f[n];
printf("%I64d\n", ans);
}
return 0;
}


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