主要思路
容易想出状态\(dp[i][j][k]\)代表前\(i\)个人在一号队列排\(j\)分钟及在二号队排\(k\)分钟的最小时间。然后发现\(k\)这一维可以直接消去，因为有\(k=sum[i]-j\)。所以把线性依赖消去后，方程就非常小清新了：
\[dp[i][j] = min\{dp[i][j], max\{dp[i – 1][j – persons[i].a], j + persons[i].b\}\},当persons[i].a\leq j\]
\[dp[i][j] = min\{dp[i][j], max\{dp[i – 1][j], sum[i] – j + persons[i].b\}\}\]
代码
// P2577.cpp
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define ll long long
using namespace std;
const int MX_N = 220;
struct person
{
ll a, b;
bool operator<(const person &ps) const
{
return b > ps.b;
}
} persons[MX_N];
ll sum[MX_N], n, dp[MX_N][MX_N * MX_N];
int main()
{
scanf("%lld", &n);
for (int i = 1; i <= n; i++)
scanf("%lld%lld", &persons[i].a, &persons[i].b);
sort(persons + 1, persons + 1 + n);
for (int i = 1; i <= n; i++)
sum[i] = sum[i - 1] + persons[i].a;
memset(dp, 0x3f, sizeof(dp));
dp[0][0] = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= sum[i]; j++)
{
if (j >= persons[i].a)
dp[i][j] = min(dp[i][j], max(dp[i - 1][j - persons[i].a], (ll)(j + persons[i].b)));
dp[i][j] = min(dp[i][j], max(dp[i - 1][j], sum[i] - j + persons[i].b));
}
ll ans = 0x7fffffff;
for (int i = 0; i <= sum[n]; i++)
ans = min(ans, dp[n][i]);
printf("%lld", ans);
return 0;
}