洛谷P2577 [ZJOI2005]午餐 打饭时间作为容量DP

P2577 [ZJOI2005]午餐

）逼着自己做DP

题意：

　　有n个人打饭，每个人都有打饭时间和吃饭时间。有两个打饭窗口，问如何安排可以使得总用时最少。

思路：

　　1）可以发现吃饭时间最长的要先打饭。（我也是看别人题解才知道）

　　2）然后就是对于前i个人，他不是在一号窗口打饭，就是在二号窗口打饭。所以用$dp[i][j]$表示前i个人，在一号窗口打饭j时间的总用时。因为$dp[i][k = sum - j] $就表示前i个人在二号窗口用时k的总用时。

 

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC                      \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define FT(A, B, C) for (int A = B; A <= C; ++A) //用来压行
#define REP(i, j, k) for (int i = j; i < k; ++i)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

/*-----------------------showtime----------------------*/

const int maxn = 209;
int dp[maxn][maxn * maxn], sum[maxn];
struct node {
    int t, w;
} a[maxn];
bool cmp(node a, node b) {
    return a.w > b.w;
}

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &a[i].t, &a[i].w);
    }
    sort(a + 1, a + 1 + n, cmp);
    for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i].t;
    memset(dp, inf, sizeof(dp));
    dp[0][0] = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = sum[i]; j >= 0; j--) {
            if (j >= a[i].t) dp[i][j] = min(dp[i][j], max(dp[i - 1][j - a[i].t], j + a[i].w));
            dp[i][j] = min(dp[i][j], max(dp[i - 1][j], sum[i] - j + a[i].w));
        }
    }
    int ans = inf;
    for (int i = 0; i <= sum[n]; i++) ans = min(ans, dp[n][i]);
    printf("%d\n", ans);
    return 0;
}