题解 ZYB和售货机
- 每个点只有一条入边/出边，会形成一棵基环树
首先考虑对选了会产生收益的点之间连边
然后发现每个连通块里有且仅有一个环（自环也算）
考场上想分情况讨论，对每个环上的点维护指向它的最大值和次大值，再维护个 \(min_{dlt}\)
然而思路不怎么清楚，没调出来
正解和暴力很像，但思路要清晰地多
对于这个环，有两种可能
若存在环边不是最优决策，那环上的点一定可以被取完，当做树处理即可
若每个环边都是最优决策，需要在环上找一个剩下的点
显然要找环边收益-连向它的树边收益最小的点（即损失最小的点）
减去这个差值即可
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
#define reg register int
#define fir first
#define sec second
#define make make_pair
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n;
int f[N], c[N], d[N], a[N];
namespace force{
int *****dp, ans;
inline void tran(int u, int au, int v, int i, int j, int k, int h, int l) {
if (!au) return ;
if (v==1) {
if (!i) return ;
dp[i-1][j][k][h][l] = max(dp[i-1][j][k][h][l], dp[i][j][k][h][l]+d[v]-c[u]);
}
else if (v==2) {
if (!j) return ;
dp[i][j-1][k][h][l] = max(dp[i][j-1][k][h][l], dp[i][j][k][h][l]+d[v]-c[u]);
}
else if (v==3) {
if (!k) return ;
dp[i][j][k-1][h][l] = max(dp[i][j][k-1][h][l], dp[i][j][k][h][l]+d[v]-c[u]);
}
else if (v==4) {
if (!h) return ;
dp[i][j][k][h-1][l] = max(dp[i][j][k][h-1][l], dp[i][j][k][h][l]+d[v]-c[u]);
}
else if (v==5) {
if (!l) return ;
dp[i][j][k][h][l-1] = max(dp[i][j][k][h][l-1], dp[i][j][k][h][l]+d[v]-c[u]);
}
}
void solve() {
dp=new int****[a[1]+1];
for (int i=0; i<=a[1]; ++i) {
dp[i]=new int***[a[2]+1];
for (int j=0; j<=a[2]; ++j) {
dp[i][j]=new int**[a[3]+1];
for (int k=0; k<=a[3]; ++k) {
dp[i][j][k]=new int*[a[4]+1];
for (int h=0; h<=a[4]; ++h)
dp[i][j][k][h]=new int[a[5]+1];
}
}
}
dp[a[1]][a[2]][a[3]][a[4]][a[5]]=0;
for (int i=a[1]; ~i; --i) {
for (int j=a[2]; ~j; --j) {
for (int k=a[3]; ~k; --k) {
for (int h=a[4]; ~h; --h) {
for (int l=a[5]; ~l; --l) {
tran(1, i, f[1], i, j, k, h, l);
tran(2, j, f[2], i, j, k, h, l);
tran(3, k, f[3], i, j, k, h, l);
tran(4, h, f[4], i, j, k, h, l);
tran(5, l, f[5], i, j, k, h, l);
ans=max(ans, dp[i][j][k][h][l]);
}
}
}
}
}
printf("%d\n", ans);
exit(0);
}
}
namespace task1{
ll ans;
void solve() {
for (int i=1; i<=n; ++i) {
if (d[i]>c[i])
ans += 1ll*(d[i]-c[i])*a[i];
}
printf("%lld\n", ans);
exit(0);
}
}
namespace task2{
ll ans, max_redlt[N];
int cnt[N], fa[N], siz[N];
pair<int, int> maxinc[N], secinc[N];
bool vis[N], vist[N];
inline int find(int p) {return fa[p]==p?p:fa[p]=find(p);}
void solve() {
for (int i=1; i<=n; ++i) fa[i]=i, siz[i]=1;
for (int i=1; i<=n; ++i) maxinc[i].fir=-INF, secinc[i].fir=-INF;
for (int i=1; i<=n; ++i) {
if (d[i]-c[i] >= maxinc[f[i]].fir) {
secinc[f[i]]=maxinc[f[i]];
maxinc[f[i]]=make(d[i]-c[i], i);
}
else if (d[i]-c[i] >= secinc[f[i]].fir) {
secinc[f[i]]=make(d[i]-c[i], i);
}
}
for (int i=1; i<=n; ++i) if (maxinc[i].fir>0) {
vis[i]=1;
ans+=1ll*maxinc[i].fir*a[i];
}
for (int i=1; i<=n; ++i) {
if (vis[i] && vis[f[i]]) {
int f1=find(i), f2=find(f[i]);
fa[f1]=f2; siz[f2]+=siz[f1];
}
}
printf("%lld\n", ans);
exit(0);
}
}
namespace task{
int head[N], size, sta[N], top, fa[N];
ll sum[N], ans, max_inc[N];
bool vis[N], ring[N];
struct egde{int to, next, val; bool back;}e[N<<1];
inline void add(int s, int t, int w, bool typ) {e[++size].to=t; e[size].next=head[s]; e[size].val=w; e[size].back=typ; head[s]=size;}
inline int find(int p) {return fa[p]==p?p:fa[p]=find(fa[p]);}
int dfs(int u) {
vis[u]=1;
for (int i=head[u],v; ~i; i=e[i].next) if (!e[i].back) {
v = e[i].to;
if (u==v) {sta[++top]=u; ring[u]=1; return -1;}
if (vis[v]) {sta[++top]=u; ring[u]=1; return v;}
int tem=dfs(v);
if (~tem) {
sta[++top]=u; ring[u]=1;
return tem==u?-1:tem;
}
break;
}
return -1;
}
void solve() {
memset(head, -1, sizeof(head));
for (int i=1; i<=n; ++i) fa[i]=i;
for (int i=1,f1,f2; i<=n; ++i)
if (c[i]<d[f[i]]) {
add(i, f[i], d[f[i]]-c[i], 0);
add(f[i], i, d[f[i]]-c[i], 1);
f1=find(i); f2=find(f[i]);
if (f1!=f2) {fa[f1]=f2; sum[f2]+=sum[f1];}
if (1ll*(d[f[i]]-c[i])*a[f[i]] > max_inc[f[i]]) {
sum[f2]+=1ll*(d[f[i]]-c[i])*a[f[i]]-max_inc[f[i]];
max_inc[f[i]]=1ll*(d[f[i]]-c[i])*a[f[i]];
}
}
for (int i=1,f1; i<=n; ++i) if (!vis[f1=find(i)]) {
//cout<<"sum: "<<sum[f1]<<endl;
dfs(i);
//cout<<"top: "<<top<<endl;
if (top<=1) ans+=sum[f1];
else {
ll min_dlt=INF, t;
for (int j=1,u; j<=top; ++j) {
u=sta[j]; t=d[f[u]]-c[u];
min_dlt=min(min_dlt, t);
for (int k=head[f[u]],v; ~k; k=e[k].next) if (e[k].back) {
v = e[k].to;
if (!ring[v]) min_dlt=min(min_dlt, t-d[f[u]]+c[v]);
}
}
//cout<<"min_dlt: "<<min_dlt<<endl;
min_dlt=max(min_dlt, 0ll);
ans+=sum[f1]-min_dlt;
}
while (top) ring[sta[top--]]=0;
vis[f1]=1;
}
printf("%lld\n", ans);
exit(0);
}
}
signed main()
{
freopen("goods.in", "r", stdin);
freopen("goods.out", "w", stdout);
n=read();
bool all_same=1;
for (int i=1; i<=n; ++i) {
f[i]=read(), c[i]=read(), d[i]=read(), a[i]=read();
if (f[i]!=i) all_same=0;
}
//if (all_same) task1::solve();
//else if (n<=5) force::solve();
//else task1::solve();
task::solve();
return 0;
}
