hdu 1024 Max Sum Plus Plus 最大m个子序列

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1024

题意

最大m个子序列

dp

dp[divid][now] = max(dp[divid-1][1~now-1], dp[divid-1][now-1])+array[now];

解释:

now 是选择第 now 个, divid 是前 now 个分为 divid 个序列.

数据范围 1e6

dp[1e6][1e6] == mle

降维

降维真不是人干的活2333333

发现每一个元素只用了几次, 有贡献的永远是最大值, 所以开个tmax记录当前divid的max, 滚动下

不清楚为啥divid = 1 的时候会wa, 大概是写挫了, 懒得改于是divid = 1 特判好了

#include <stdio.h>
#include <iostream>
#include <cstring>
using namespace std;
const int MAX = 1e6+10;
const long long INF = 1e15+10;
long long dp[MAX], ary[MAX], n, m, itor, tmax, nxt, last, ans;
long long mmax (long long a, long long b) {
return a > b ? a : b;
}
bool getin() {
if (scanf("%lld%lld", &m, &n) == 2) {
for (int i = 0; i < n; ++i)
scanf("%lld", ary+i);
memset (dp, 0, sizeof(dp));
return true;
}
return false;
}
void judge() {
for (int divid = 1; divid <= m; ++divid) {
for (int now = divid-1; now < n; ++now) {
if (divid == 1) {
if (!now)
tmax = mmax(0, ary[0]), dp[now] = ary[0];
else
dp[now] = mmax(dp[now-1]+ary[now], ary[now]);
}
else {
if (now == divid-1)
tmax = dp[now-1];
nxt = mmax(tmax, dp[now]);
dp[now] = mmax(tmax, dp[now-1])+ary[now];
tmax = nxt;
}
//          printf("%cdp[%d] = %lld", "\n "[(bool)(now-divid+1)], now, dp[now]);
}
}
ans = -INF;
for (int i = m-1; i < n; ++i)
ans = mmax(ans, dp[i]);
printf("%lld\n", ans);
}
int main() {
freopen("in.t", "r", stdin);
freopen("out.t", "w", stdout);
while( getin()) {
judge();
}
return 0;
}

2017-09-06