cf1271 D. Portals

Link

Link

Solution

如果最后的答案中我决定要占领 $i$，那么所有能占领 $i$的位置中我选最后一个肯定最好

因为如果 $i$到 $x$之间可能会增加兵力，这样做能够让其对其他选择的影响尽量小

有了这条性质，就可以直接 $dp_{ij}$表示打下了前 $i$座城堡，并且还剩下 $j$的兵力的最大获利

Code

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
ll a[maxn], b[maxn], c[maxn], n, K, fa[maxn], dp[5050][5050];
vector<ll> e[maxn];
int main()
{
    ll i, j, k, m, u, v;
    n=read(), m=read(), K=read();
    rep(i,1,n)a[i]=read(), b[i]=read(), c[i]=read();
    rep(i,1,m)u=read(), v=read(), fa[v]=max(fa[v],u);
    rep(i,1,n)fa[i]=max(i,fa[i]);
    rep(i,1,n)e[fa[i]].emb(i);
    rep(i,0,n)rep(j,0,5000)dp[i][j]=-linf;
    dp[0][K]=0;
    rep(i,0,n)
    {
        sort(e[i+1].begin(),e[i+1].end(),[&](ll x, ll y){return c[x]>c[y];});
        rep(j,0,5000)
        {
            if(dp[i][j]<0)continue;
            // printf("dp[%lld][%lld]=%lld\n",i,j,dp[i][j]);
            if(j<a[i+1])continue;
            if(j>=a[i+1] and j+b[i+1]<=5000)dp[i+1][j+b[i+1]]=max(dp[i][j],dp[i+1][j+b[i+1]]);
            ll s=0;
            rep(k,1,min((ll)e[i+1].size(),j+b[i+1]))
            {
                s+=c[e[i+1].at(k-1)];
                auto &to=dp[i+1][j+b[i+1]-k];
                to=max(to,dp[i][j]+s);
            }
        }
    }
    ll ans=-1;
    rep(j,0,5000)ans=max(ans,dp[n][j]);
    cout<<ans;
    return 0;
}