1. CF730I Olympiad in Programming and Sports
大意: $n$个人, 第$i$个人编程能力$a_i$, 运动能力$b_i$, 要选出$p$个组成编程队, $s$个组成运动队, 每个队的收益为队员能力和, 求最大收益.
费用流做法很显然, 开两个点$X,Y$表示编程和运动, 源点向每个人连边, 代价为$0$, 每个人向$X$连边, 代价为编程能力, 每个人向$Y$连边, 代价为运动能力, $X$向汇点连边容量为$p$, $Y$向汇点连边, 容量为$s$, 然后求出最大费用最大流即为答案.
考虑用堆模拟费用流. 每次增广只有四种情况.
- 添加一个未被选择的人去编程队
- 添加一个未被选择的人去运动队
- 让一个人从运动队到编程队, 再选择一个未被选择的人去运动队
- 让一个人从编程队到运动队, 再选择一个未被选择的人去编程队
前两种情况直接用堆维护最大即可. 后两种情况的话, 在每次决策时, 都把替换的收益扔进一个堆里即可.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+10; int n, x, y; int a[N], b[N], va[N], vb[N]; pii s[5]; priority_queue<pii> q[5]; ll ans; pii get(priority_queue<pii> &q, int *v1, int *v2) { while (q.size()&&(v1[q.top().y]||v2[q.top().y])) q.pop(); return q.empty()?pii(-INF,0):q.top(); } void add(int u, int tp) { if (va[u]) ans-=a[u],va[u]=0,++x; if (vb[u]) ans-=b[u],vb[u]=0,++y; if (tp==1) ans+=a[u],va[u]=1,--x,q[4].push(pii(b[u]-a[u],u)); else ans+=b[u],vb[u]=1,--y,q[3].push(pii(a[u]-b[u],u)); } int main() { scanf("%d%d%d", &n, &x, &y); REP(i,1,n) scanf("%d",a+i); REP(i,1,n) scanf("%d",b+i); REP(i,1,n) { q[1].push(pii(a[i],i)); q[2].push(pii(b[i],i)); } int tot = x+y; REP(i,1,tot) { s[1] = get(q[1],va,vb); s[2] = get(q[2],va,vb); s[3] = get(q[3],va,va); s[4] = get(q[4],vb,vb); s[3].x += s[2].x; s[4].x += s[1].x; if (!x) s[1].x=s[3].x=-INF; if (!y) s[2].x=s[4].x=-INF; int p = max_element(s+1,s+5)-s; if (p==1) add(s[1].y,1); else if (p==2) add(s[2].y,2); else if (p==3) add(s[3].y,1),add(s[2].y,2); else add(s[4].y,2),add(s[1].y,1); } printf("%lld ", ans); REP(i,1,n) if (va[i]) printf("%d ",i);hr; REP(i,1,n) if (vb[i]) printf("%d ",i);hr; }
2. NOI2019 序列
大意: 给定两个序列$a,b$, 要求每个序列选出$K$个元素, 且位置相同的元素要至少选$L$个, 求选出元素的最大和.
首先可以得到一个显然的费用流做法.
源点向$a$中每个元素连边, 代价为$a_i$, $b$中每个元素向汇点连边, 代价为$b_i$.
$a$中每个点向对应位置的$b$中的点连边.
再开两个点$X,Y$, $X$向$Y$连容量为$K-L$的边, 表示可以选出$K-L$个不一样的位置.
$a$中每个点再连向$X$, $Y$连向$b$中每个点.
最后求出源点到汇点容量为$K$的最大费用流即为答案.
考虑用堆模拟费用流, 分四种情况.
- 从$a$中选最大,$b$中也选最大 (要求X到Y仍有剩余流量)
- $a$,$b$选出位置相同
- $a$中选最大,$b$中选出之前选过的$a$对应位置
- $b$中选最大,$a$中选出之前选过的$b$对应位置
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head #ifdef ONLINE_JUDGE const int N = 1e6+10; #else const int N = 567; #endif int n, k, l, a[N], b[N]; int va[N], vb[N]; priority_queue<pii> q[10]; struct t3 {int x,a,b;} f[10]; ll ans; pii s[10]; int cost(int x, int y) { int ret = 0; if (x!=y) { if (vb[x]) ++ret; else --ret; if (va[y]) ++ret; else --ret; } return ret<0?-1:ret>0?1:0; } void add(int x, int y) { l += cost(x,y); ans += a[x]+b[y]; va[x] = vb[y] = 1; if (!vb[x]) q[4].push(pii(b[x],x)); if (!va[y]) q[3].push(pii(a[y],y)); } pii get(priority_queue<pii> &q, int *v1, int *v2) { while (q.size()&&(v1[q.top().y]||v2[q.top().y])) q.pop(); return q.empty()?pii(-INF,0):q.top(); } int main() { int t; scanf("%d", &t); while (t--) { REP(i,0,4) while (q[i].size()) q[i].pop(); scanf("%d%d%d", &n, &k, &l); REP(i,1,n) scanf("%d",a+i); REP(i,1,n) scanf("%d",b+i); REP(i,1,n) { va[i] = vb[i] = 0; q[0].push(pii(a[i]+b[i],i)); q[1].push(pii(a[i],i)); q[2].push(pii(b[i],i)); } l = k-l, ans = 0; REP(i,1,k) { s[0] = get(q[0],va,vb); s[1] = get(q[1],va,va); s[2] = get(q[2],vb,vb); s[3] = get(q[3],va,va); s[4] = get(q[4],vb,vb); f[0] = {s[1].x+s[2].x,s[1].y,s[2].y}; f[1] = {s[0].x,s[0].y,s[0].y}; f[2] = {s[1].x+s[4].x,s[1].y,s[4].y}; f[3] = {s[2].x+s[3].x,s[3].y,s[2].y}; if (l+cost(f[0].a,f[0].b)<0) f[0].x=-INF; int p = max_element(f,f+4,[](t3 a, t3 b){return a.x<b.x;})-f; add(f[p].a,f[p].b); } printf("%lld ", ans); } }
3. CF802O April Fools' Problem (hard)
大意: $n$道题, 第$i$天可以花费$a_i$准备一道题, 花费$b_i$打印一道题, 每天最多准备一道, 最多打印一道, 准备的题可以留到以后打印, 求最少花费使得准备并打印$k$道题.
这个题好有意思, 直接用堆很难模拟, 似乎可以用线段树来模拟. 一个非常好写的做法是带权二分,带权二分具体原理其实一直都没搞懂。。。
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+50; int n, k, a[N], b[N]; priority_queue<int,vector<int>,greater<int> > q[2]; int main() { scanf("%d%d", &n, &k); REP(i,1,n) scanf("%d", a+i); REP(i,1,n) scanf("%d", b+i); int l = 0, r = 2e9; ll ans = 0; while (l<=r) { REP(i,0,1) while (q[i].size()) q[i].pop(); ll ret = 0; int cnt = 0, mid = (ll)l+r>>1; REP(i,1,n) { q[0].push(a[i]); ll t1 = q[0].top()+(ll)b[i]-mid; ll t2 = q[1].empty()?INF:b[i]+q[1].top(); if (t1<=0&&t1<=t2) { ret += t1, ++cnt; q[1].push(-b[i]); q[0].pop(); } else if (t2<0) { ret += t2; q[1].pop(); q[1].push(-b[i]); } } if (cnt>=k) ans=ret+(ll)k*mid,r=mid-1; else l=mid+1; } printf("%lld ", ans); }
4. hdu 6698 Coins
大意: $n$组硬币, 第$i$组有一个$a_i$元的和$b_i$元的, 每组要么不选, 要么选$a_i$, 要么全选, 对于$kle 2n$, 求出拿$k$个硬币的最大钱数.
堆模拟这四种方案即可:
- 取一个最大的$a$
- 补上一个最大的$b$
- 删掉一个$a$, 取一个最大的$a+b$
- 删掉一个$b$, 取一个最大的$a+b$
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ' ' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head const int N = 1e6+10; int n, a[N], b[N]; int va[N], vb[N]; priority_queue<pii> q[3]; priority_queue<pii,vector<pii>,greater<pii> > qq[2]; pii s[4]; void add(int u, int tp) { if (tp==0) va[u]=1,q[1].push(pii(b[u],u)),qq[0].push(pii(a[u],u)); else vb[u]=1,qq[1].push(pii(b[u],u)); } void work() { scanf("%d", &n); REP(i,0,2) while (q[i].size()) q[i].pop(); REP(i,0,1) while (qq[i].size()) qq[i].pop(); REP(i,1,n) { scanf("%d%d",a+i,b+i); q[0].push(pii(a[i],i)); q[2].push(pii(a[i]+b[i],i)); va[i] = vb[i] = 0; } ll ans = 0; REP(i,1,2*n) { while (q[0].size()&&va[q[0].top().y]) q[0].pop(); s[0] = q[0].empty()?pii(-INF,0):q[0].top(); while (q[1].size()&&(!va[q[1].top().y]||vb[q[1].top().y])) q[1].pop(); s[1] = q[1].empty()?pii(-INF,0):q[1].top(); while (q[2].size()&&va[q[2].top().y]) q[2].pop(); s[2] = s[3] = q[2].empty()?pii(-INF,0):q[2].top(); if (i==1) s[2].x = s[3].x = -INF; else { while (qq[0].size()&&(!va[qq[0].top().y]||vb[qq[0].top().y])) qq[0].pop(); if (qq[0].size()) s[2].x -= qq[0].top().x; else s[2].x = -INF; while (qq[1].size()&&!vb[qq[1].top().y]) qq[1].pop(); if (qq[1].size()) s[3].x -= qq[1].top().x; else s[3].x = -INF; } int p = max_element(s,s+4)-s; ans += s[p].x; if (p<=1) add(s[p].y,p); else { //要注意先删除在添加, 不然qq会变 if (p==2) va[qq[0].top().y]=0,q[0].push(qq[0].top()); else vb[qq[1].top().y]=0,q[1].push(qq[1].top()); add(s[p].y,0),add(s[p].y,1); } printf("%lld%c",ans," "[i==2*n]); } } int main() { int t; scanf("%d", &t); while (t--) work(); }