CCF CSP 2020-06-05 乔乔和牛牛逛超市
买了x才能买y 等价于要想得到y的最大收益就必须连带着把x的最大收益(即使可能<0)也得收下，这个模型是典型的最大权闭合子图，答案应该是图中正的点权和-最小割的容量。最小割跑最大流模板就可得到。但是第二类点的建图方式还没想到，可能是需要拆点。
80分代码（边建反了） 就这也能拿80分？ ccf数据好水啊
#include
#include
#include
#include
#define LL long long
using namespace std;
const int N = 100010, M = 2000010;
const LL INF = 2e9;
int n,m,S,T;
int h[N],tot;
struct Edge{
int to,nxt;
LL f;
}e[M];
int q[N],d[N],cur[N],w[N];
void add(int a,int b,LL c)
{
e[tot].to = b, e[tot].f = c, e[tot].nxt = h[a], h[a] = tot ++;
e[tot].to = a, e[tot].f = 0, e[tot].nxt = h[b], h[b] = tot ++;
}
bool bfs()
{
memset(d, -1, sizeof d);
int hh = 0, tt = -1;
q[++ tt] = S; d[S] = 0; cur[S] = h[S];
while(hh <= tt)
{
int u = q[hh ++];
for(int i = h[u]; ~i; i = e[i].nxt)
{
int to = e[i].to;
if(d[to] == -1 && e[i].f)
{
d[to] = d[u] + 1;
cur[to] = h[to];
q[++ tt] = to;
if(to == T) return true;
}
}
}
return false;
}
LL dfs(int u,int limit)
{
if(u == T) return limit;
LL flow = 0;
for(int i = cur[u]; ~i && flow < limit; i = e[i].nxt)
{
cur[u] = i;
int to = e[i].to;
if(d[to] == d[u] + 1 && e[i].f)
{
LL f = dfs(to, min(e[i].f, limit - flow));
if(!f) d[to] = -1;
e[i].f -= f, e[i ^ 1].f += f, flow += f;
}
}
return flow;
}
LL dinic()
{
LL r = 0, flow;
while(bfs()) while(flow = dfs(S,INF)) r += flow;
return r;
}
int gety(int a, int b, int c,int x)
{
return a * x * x + b * x + c;
}
int cal(int a,int b, int c, int l, int r)
{
int w = -INF;
if(a > 0) w = max(gety(a, b, c, l), gety(a, b, c, r));
else if(a < 0)
{
int p = round(-(double)b / (2 * a));
if(p < l) w = gety(a, b, c, l);
else if(p > r) w = gety(a, b, c, r);
else w = gety(a, b, c, p);
}
else
{
if(b > 0) w = gety(a, b, c, r);
else w = gety(a, b, c, l);
}
return w;
}
int main()
{
memset(h, -1, sizeof h);
cin >> n >> m;
S = 0, T = n * 2 + 1;
for(int i = 1; i <= n; ++i)
{
int l, r, a, b, c;
scanf("%d%d%d%d%d",&l, &r, &a, &b, &c);
w[i] = cal(a, b, c, l + 1, r - 1);
w[i + n] = max(gety(a, b, c, l),gety(a, b, c, r)) - w[i];
add(i + n, i, INF);
}
for(int i = 0; i < m; ++i)
{
int x, y, z; scanf("%d%d%d",&z, &x, &y);
if(z == 1) add(x, y, INF);
else add(x + n, y, INF); // 2e5
}
LL sum = 0;
for(int i = 1; i <= n * 2; ++i)
{
//cout << w[i] << ' ' << i << '\n';
if(w[i] < 0) add(i, T, -w[i]);
else if(w[i] > 0) add(S, i, w[i]), sum += w[i];
}
cout << sum - dinic();
} 100分代码：(swap一下输入的边)
#include
#include
#include
#include
#define LL long long
using namespace std;
const int N = 100010, M = 2000010;
const LL INF = 2e9;
int n,m,S,T;
int h[N],tot;
struct Edge{
int to,nxt;
LL f;
}e[M];
int q[N],d[N],cur[N],w[N];
void add(int a,int b,LL c)
{
e[tot].to = b, e[tot].f = c, e[tot].nxt = h[a], h[a] = tot ++;
e[tot].to = a, e[tot].f = 0, e[tot].nxt = h[b], h[b] = tot ++;
}
bool bfs()
{
memset(d, -1, sizeof d);
int hh = 0, tt = -1;
q[++ tt] = S; d[S] = 0; cur[S] = h[S];
while(hh <= tt)
{
int u = q[hh ++];
for(int i = h[u]; ~i; i = e[i].nxt)
{
int to = e[i].to;
if(d[to] == -1 && e[i].f)
{
d[to] = d[u] + 1;
cur[to] = h[to];
q[++ tt] = to;
if(to == T) return true;
}
}
}
return false;
}
LL dfs(int u,int limit)
{
if(u == T) return limit;
LL flow = 0;
for(int i = cur[u]; ~i && flow < limit; i = e[i].nxt)
{
cur[u] = i;
int to = e[i].to;
if(d[to] == d[u] + 1 && e[i].f)
{
LL f = dfs(to, min(e[i].f, limit - flow));
if(!f) d[to] = -1;
e[i].f -= f, e[i ^ 1].f += f, flow += f;
}
}
return flow;
}
LL dinic()
{
LL r = 0, flow;
while(bfs()) while(flow = dfs(S,INF)) r += flow;
return r;
}
int gety(int a, int b, int c,int x)
{
return a * x * x + b * x + c;
}
int cal(int a,int b, int c, int l, int r)
{
int w = -INF;
if(a > 0) w = max(gety(a, b, c, l), gety(a, b, c, r));
else if(a < 0)
{
int p = round(-(double)b / (2 * a));
if(p < l) w = gety(a, b, c, l);
else if(p > r) w = gety(a, b, c, r);
else w = gety(a, b, c, p);
}
else
{
if(b > 0) w = gety(a, b, c, r);
else w = gety(a, b, c, l);
}
return w;
}
int main()
{
memset(h, -1, sizeof h);
cin >> n >> m;
S = 0, T = n * 2 + 1;
for(int i = 1; i <= n; ++i)
{
int l, r, a, b, c;
scanf("%d%d%d%d%d",&l, &r, &a, &b, &c);
w[i] = cal(a, b, c, l + 1, r - 1);
w[i + n] = max(gety(a, b, c, l),gety(a, b, c, r)) - w[i];
add(i + n, i, INF);
}
for(int i = 0; i < m; ++i)
{
int x, y, z; scanf("%d%d%d",&z, &x, &y);
swap(x, y);
if(z == 1) add(x, y, INF);
else add(x + n, y, INF); // 2e5
}
LL sum = 0;
for(int i = 1; i <= n * 2; ++i)
{
//cout << w[i] << ' ' << i << '\n';
if(w[i] < 0) add(i, T, -w[i]);
else if(w[i] > 0) add(S, i, w[i]), sum += w[i];
}
cout << sum - dinic();
}