Codeforces Round #579 (Div. 3) Complete the Projects（贪心、DP）

http://codeforces.com/contest/1203/problem/F1

Examples

input 1

 -
 -

output 1

YES

input 2

 -
 -
 

output 2

YES

input 3

 -

 -

output 3

YES

input 4

 -
 

output 4

NO

Note

In the first example, the possible order is: 1,2,3.

In the second example, the possible order is: 2,3,1.

In the third example, the possible order is: 3,1,4,2.

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <string>
 #include <math.h>
 #include <algorithm>
 #include <queue>
 #include <set>
 #include <math.h>
 const int INF=0x3f3f3f3f;
 typedef long long LL;
 const int mod=1e9+;
 const double PI=acos(-);
 const int maxn=1e5+;
 using namespace std;

 struct node1
 {
     int a;
     int b;
 }zheng[];

 struct node2
 {
     int a;
     int b;
 }fu[];

 int n1,n2;

 bool cmp1(node1 x,node1 y)
 {
     return x.a<y.a;
 }

 bool cmp2(node2 x,node2 y)
 {
     if(x.a+x.b!=y.a+y.b)
         return x.a+x.b>y.a+y.b;
     else
         return x.b>y.b;
 }

 int main()
 {
     int n,r;
     scanf("%d %d",&n,&r);
     for(int i=;i<=n;i++)
     {
         int x,y;
         scanf("%d %d",&x,&y);
         if(y<)
         {
             fu[n2].a=x;
             fu[n2].b=y;
             n2++;
         }
         else
         {
             zheng[n1].a=x;
             zheng[n1].b=y;
             n1++;
         }
     }
     sort(zheng,zheng+n1,cmp1);
     for(int i=;i<n1;i++)
     {
         if(r<zheng[i].a)
         {
             printf("NO\n");
             return ;
         }
         r+=zheng[i].b;
     }
     sort(fu,fu+n2,cmp2);
     for(int i=;i<n2;i++)
     {
         if(r<fu[i].a)
         {
             printf("NO\n");
             return ;
         }
         r+=fu[i].b;
     }
     if(r<)
         printf("NO\n");
     else
         printf("YES\n");
     return ;
 }

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <queue>
 using namespace std;
 #define endl '\n'

 const int maxn = ;
 int dp[];
 pair<int, int> a[maxn];
 queue<int> q;

 int main(){
     ios::sync_with_stdio(false);
     cin.tie(NULL);

     int n, r;
     cin >> n >> r;

     for(int i = ; i < n; i++){
         cin >> a[i].first >> a[i].second;
     } 

     sort(a, a + n);

     int l = , sum = r, ans = ;
     for(; l < n && a[l].first <= sum; l++){
         if(a[l].second >= ){
             q.push(l);
             ans++;
         }
     }

     while(!q.empty()){
         int c = q.front();
         q.pop();

         sum += a[c].second;

         for(; l < n && a[l].first <= sum; l++){
             if(a[l].second >= ){
                 q.push(l);
                 ans++;
             }
         }
     }

     sort(a, a + l, [&](pair<int, int> f, pair<int, int> g){
         return f.first + f.second > g.first + g.second;
     });
     dp[] = ans;

     for(int i = ; i < l; i++){
         if(a[i].second >= ) continue;
         int y = -a[i].second;

         for(int j = sum - max(y, a[i].first); j >= ; j--){
             dp[j + y] = max(dp[j + y], dp[j] + );
             ans = max(ans, dp[j + y]);
         }
     }

     cout << (ans == n ? "YES" : "NO") << endl;

     return ;
 }

---

http://codeforces.com/contest/1203/problem/F2

Examples

input 1

 -
 -

output 1

input 2

 -
 -

 -

output 2

input 3

 -

output 3

先粘题解，以后再填坑

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <string>
 #include <math.h>
 #include <algorithm>
 #include <queue>
 #include<bits/stdc++.h>
 using namespace std;

 #define PI acos(-1)
 #define hell 1000000007
 #define HELL 998244353
 #define io ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
 #define fix(n) cout << fixed << setprecision(n)
 #define mset(a,n) memset(a,n,sizeof a)
 #define rep(i,a,b) for (__typeof((b)) i=(a);i<(b);i++)
 #define repp(i,a,b,p) for(__typeof((b)) i=(a);i<(b);i+=p)
 #define ren(i,a,b) for(__typeof((a)) i=(a);i>=(b);i--)
 #define renn(i,a,b,p) for(__typeof((a) i=(a);i>=(b);i-=p)
 #define ADD(a,b,c) ((a)%c+(b)%c)%c
 #define SUB(a,b,c) ((a)%c-(b)%c+c)%c
 #define MUL(a,b,c) (((a)%c)*((b)%c))%c
 #define lbd lower_bound
 #define ubd upper_bound
 #define ll long long
 #define ld long double
 #define pb push_back
 #define fi first
 #define se second
 #define vll vector<ll>
 #define pll pair<ll,ll>
 #define vpll vector<pll>
 #define all(v) (v).begin(), (v).end()
 #define sz(x) (ll)x.size()
 #define endl "\n"
 #define out(n) cout<<n<<" "
 #define outl(n) cout<<n<<endl
 #define line cout<<endl
 #define bug(n) {outl(n);return;}
 #define N 105
 ll n,r,dp[N][];
 pll a[N];
 bool comp(pll a, pll b){
     if(a.se>&&b.se>)return a.fi<b.fi;
     if(a.se>||b.se>)return a.se>;
     return a.fi+a.se>b.fi+b.se;
 }
 ll fun(ll i, ll r){
     if(dp[i][r]!=-)return dp[i][r];
     if(i==n+)return ;
     ll ans=fun(i+,r);
     if(r>=a[i].fi&&r>=-a[i].se)ans=max(ans,+fun(i+,r+a[i].se));
     return dp[i][r]=ans;
 }
 void solve(){
     cin>>n>>r;
     mset(dp,-);
     rep(i,,n+)cin>>a[i].fi>>a[i].se;
     sort(a+,a+n+,comp);
     bug(fun(,r));
 }
 void prep(){

 }
 int main(){
     io;
     ll t=;
     // cin>>t;
     prep();
     fix();
     while(t--)
         solve();
     return ;
 }

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <queue>
 using namespace std;
 #define endl '\n'

 const int maxn = ;
 int dp[];
 pair<int, int> a[maxn];
 queue<int> q;

 int main(){
     ios::sync_with_stdio(false);
     cin.tie(NULL);

     int n, r;
     cin >> n >> r;

     for(int i = ; i < n; i++){
         cin >> a[i].first >> a[i].second;
     } 

     sort(a, a + n);

     int l = , sum = r, ans = ;
     for(; l < n && a[l].first <= sum; l++){
         if(a[l].second >= ){
             q.push(l);
             ans++;
         }
     }

     while(!q.empty()){
         int c = q.front();
         q.pop();

         sum += a[c].second;

         for(; l < n && a[l].first <= sum; l++){
             if(a[l].second >= ){
                 q.push(l);
                 ans++;
             }
         }
     }

     sort(a, a + l, [&](pair<int, int> f, pair<int, int> g){
         return f.first + f.second > g.first + g.second;
     });
     dp[] = ans;

     for(int i = ; i < l; i++){
         if(a[i].second >= ) continue;
         int y = -a[i].second;

         for(int j = sum - max(y, a[i].first); j >= ; j--){
             dp[j + y] = max(dp[j + y], dp[j] + );
             ans = max(ans, dp[j + y]);
         }
     }

     cout << ans << endl;

     return ;
 }