http://codeforces.com/contest/1203/problem/F1
Examples
input 1
3 4 4 6 10 -2 8 -1
output 1
YES
input 2
3 5 4 -5 4 -2 1 3
output 2
YES
input 3
4 4 5 2 5 -3 2 1 4 -2
output 3
YES
input 4
3 10 10 0 10 -10 30 0
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.
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <string> 5 #include <math.h> 6 #include <algorithm> 7 #include <queue> 8 #include <set> 9 #include <math.h> 10 const int INF=0x3f3f3f3f; 11 typedef long long LL; 12 const int mod=1e9+7; 13 const double PI=acos(-1); 14 const int maxn=1e5+10; 15 using namespace std; 16 17 struct node1 18 { 19 int a; 20 int b; 21 }zheng[105]; 22 23 struct node2 24 { 25 int a; 26 int b; 27 }fu[105]; 28 29 int n1,n2; 30 31 bool cmp1(node1 x,node1 y) 32 { 33 return x.a<y.a; 34 } 35 36 bool cmp2(node2 x,node2 y) 37 { 38 if(x.a+x.b!=y.a+y.b) 39 return x.a+x.b>y.a+y.b; 40 else 41 return x.b>y.b; 42 } 43 44 int main() 45 { 46 int n,r; 47 scanf("%d %d",&n,&r); 48 for(int i=1;i<=n;i++) 49 { 50 int x,y; 51 scanf("%d %d",&x,&y); 52 if(y<0) 53 { 54 fu[n2].a=x; 55 fu[n2].b=y; 56 n2++; 57 } 58 else 59 { 60 zheng[n1].a=x; 61 zheng[n1].b=y; 62 n1++; 63 } 64 } 65 sort(zheng,zheng+n1,cmp1); 66 for(int i=0;i<n1;i++) 67 { 68 if(r<zheng[i].a) 69 { 70 printf("NO "); 71 return 0; 72 } 73 r+=zheng[i].b; 74 } 75 sort(fu,fu+n2,cmp2); 76 for(int i=0;i<n2;i++) 77 { 78 if(r<fu[i].a) 79 { 80 printf("NO "); 81 return 0; 82 } 83 r+=fu[i].b; 84 } 85 if(r<0) 86 printf("NO "); 87 else 88 printf("YES "); 89 return 0; 90 }
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <queue> 5 using namespace std; 6 #define endl ' ' 7 8 const int maxn = 100; 9 int dp[600001]; 10 pair<int, int> a[maxn]; 11 queue<int> q; 12 13 int main(){ 14 ios::sync_with_stdio(false); 15 cin.tie(NULL); 16 17 int n, r; 18 cin >> n >> r; 19 20 for(int i = 0; i < n; i++){ 21 cin >> a[i].first >> a[i].second; 22 } 23 24 sort(a, a + n); 25 26 int l = 0, sum = r, ans = 0; 27 for(; l < n && a[l].first <= sum; l++){ 28 if(a[l].second >= 0){ 29 q.push(l); 30 ans++; 31 } 32 } 33 34 while(!q.empty()){ 35 int c = q.front(); 36 q.pop(); 37 38 sum += a[c].second; 39 40 for(; l < n && a[l].first <= sum; l++){ 41 if(a[l].second >= 0){ 42 q.push(l); 43 ans++; 44 } 45 } 46 } 47 48 sort(a, a + l, [&](pair<int, int> f, pair<int, int> g){ 49 return f.first + f.second > g.first + g.second; 50 }); 51 dp[0] = ans; 52 53 for(int i = 0; i < l; i++){ 54 if(a[i].second >= 0) continue; 55 int y = -a[i].second; 56 57 for(int j = sum - max(y, a[i].first); j >= 0; j--){ 58 dp[j + y] = max(dp[j + y], dp[j] + 1); 59 ans = max(ans, dp[j + y]); 60 } 61 } 62 63 cout << (ans == n ? "YES" : "NO") << endl; 64 65 return 0; 66 }
http://codeforces.com/contest/1203/problem/F2
Examples
input 1
3 4 4 6 10 -2 8 -1
output 1
3
input 2
5 20 45 -6 34 -15 10 34 1 27 40 -45
output 2
5
input 3
3 2 300 -300 1 299 1 123
output 3
3
先粘题解，以后再填坑
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <string> 5 #include <math.h> 6 #include <algorithm> 7 #include <queue> 8 #include<bits/stdc++.h> 9 using namespace std; 10 11 #define PI acos(-1) 12 #define hell 1000000007 13 #define HELL 998244353 14 #define io ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0) 15 #define fix(n) cout << fixed << setprecision(n) 16 #define mset(a,n) memset(a,n,sizeof a) 17 #define rep(i,a,b) for (__typeof((b)) i=(a);i<(b);i++) 18 #define repp(i,a,b,p) for(__typeof((b)) i=(a);i<(b);i+=p) 19 #define ren(i,a,b) for(__typeof((a)) i=(a);i>=(b);i--) 20 #define renn(i,a,b,p) for(__typeof((a) i=(a);i>=(b);i-=p) 21 #define ADD(a,b,c) ((a)%c+(b)%c)%c 22 #define SUB(a,b,c) ((a)%c-(b)%c+c)%c 23 #define MUL(a,b,c) (((a)%c)*((b)%c))%c 24 #define lbd lower_bound 25 #define ubd upper_bound 26 #define ll long long 27 #define ld long double 28 #define pb push_back 29 #define fi first 30 #define se second 31 #define vll vector<ll> 32 #define pll pair<ll,ll> 33 #define vpll vector<pll> 34 #define all(v) (v).begin(), (v).end() 35 #define sz(x) (ll)x.size() 36 #define endl " " 37 #define out(n) cout<<n<<" " 38 #define outl(n) cout<<n<<endl 39 #define line cout<<endl 40 #define bug(n) {outl(n);return;} 41 #define N 105 42 ll n,r,dp[N][300005]; 43 pll a[N]; 44 bool comp(pll a, pll b){ 45 if(a.se>0&&b.se>0)return a.fi<b.fi; 46 if(a.se>0||b.se>0)return a.se>0; 47 return a.fi+a.se>b.fi+b.se; 48 } 49 ll fun(ll i, ll r){ 50 if(dp[i][r]!=-1)return dp[i][r]; 51 if(i==n+1)return 0; 52 ll ans=fun(i+1,r); 53 if(r>=a[i].fi&&r>=-a[i].se)ans=max(ans,1+fun(i+1,r+a[i].se)); 54 return dp[i][r]=ans; 55 } 56 void solve(){ 57 cin>>n>>r; 58 mset(dp,-1); 59 rep(i,1,n+1)cin>>a[i].fi>>a[i].se; 60 sort(a+1,a+n+1,comp); 61 bug(fun(1,r)); 62 } 63 void prep(){ 64 65 } 66 int main(){ 67 io; 68 ll t=1; 69 // cin>>t; 70 prep(); 71 fix(12); 72 while(t--) 73 solve(); 74 return 0; 75 }
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <queue> 5 using namespace std; 6 #define endl ' ' 7 8 const int maxn = 100; 9 int dp[600001]; 10 pair<int, int> a[maxn]; 11 queue<int> q; 12 13 int main(){ 14 ios::sync_with_stdio(false); 15 cin.tie(NULL); 16 17 int n, r; 18 cin >> n >> r; 19 20 for(int i = 0; i < n; i++){ 21 cin >> a[i].first >> a[i].second; 22 } 23 24 sort(a, a + n); 25 26 int l = 0, sum = r, ans = 0; 27 for(; l < n && a[l].first <= sum; l++){ 28 if(a[l].second >= 0){ 29 q.push(l); 30 ans++; 31 } 32 } 33 34 while(!q.empty()){ 35 int c = q.front(); 36 q.pop(); 37 38 sum += a[c].second; 39 40 for(; l < n && a[l].first <= sum; l++){ 41 if(a[l].second >= 0){ 42 q.push(l); 43 ans++; 44 } 45 } 46 } 47 48 sort(a, a + l, [&](pair<int, int> f, pair<int, int> g){ 49 return f.first + f.second > g.first + g.second; 50 }); 51 dp[0] = ans; 52 53 for(int i = 0; i < l; i++){ 54 if(a[i].second >= 0) continue; 55 int y = -a[i].second; 56 57 for(int j = sum - max(y, a[i].first); j >= 0; j--){ 58 dp[j + y] = max(dp[j + y], dp[j] + 1); 59 ans = max(ans, dp[j + y]); 60 } 61 } 62 63 cout << ans << endl; 64 65 return 0; 66 }