)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; //
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //
const int mod = 1e9+;
const double esp = 1e-;
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=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
return x=f?-x:x;
} /*-----------------------showtime----------------------*/ const int maxn = ;
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=; i<=n; i++){
scanf("%d%d", &a[i].t, &a[i].w);
}
sort(a+,a++n,cmp);
for(int i=; i<=n; i++) sum[i] = sum[i-] + a[i].t;
memset(dp, inf, sizeof(dp));
dp[][] = ;
for(int i=; i<=n; i++){
for(int j=sum[i]; j>=; j--){
if(j >= a[i].t)dp[i][j] = min(dp[i][j], max(dp[i-][j-a[i].t], j + a[i].w));
dp[i][j] = min(dp[i][j], max(dp[i-][j], sum[i] - j + a[i].w));
}
}
int ans = inf;
for(int i=; i<= sum[n]; i++) ans = min(ans, dp[n][i]);
printf("%d\n", ans);
return ;
}