題目鏈接 BZOJ 1565
看到這道題之後很容易想到的就是最大權閉合子圖了，但是卻有個問題就是要去除掉那些環，因爲構成了環之後，相當於是無敵的狀態，它們就永遠不會得到貢獻，並且環之後的點也是得不到貢獻的，所以，這裏利用拓撲，知道哪些點是可以作爲最大權閉合子圖上的點的，然後再根據最大權閉合子圖的思想來解決該問題即可。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e3 + 7, maxM = 3e5 + 7;
int N, M, head[maxN], cnt, score[maxN];
bool zihuan[maxN] = {false};
struct Eddge
{
int nex, to; ll flow;
Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), flow(c) {}
}edge[maxM];
inline void addEddge(int u, int v, ll w)
{
edge[cnt] = Eddge(head[u], v, w);
head[u] = cnt++;
}
inline void _add(int u, int v, ll w) { addEddge(u, v, w); addEddge(v, u, 0); }
struct Max_Flow
{
int deep[maxN], cur[maxN], S, T, node;
queue<int> Q;
inline bool bfs()
{
while(!Q.empty()) Q.pop();
for(int i=0; i<=node; i++) deep[i] = 0;
deep[S] = 1;
Q.push(S);
int u;
while(!Q.empty())
{
u = Q.front(); Q.pop();
ll f;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to; f = edge[i].flow;
if(f && !deep[v])
{
deep[v] = deep[u] + 1;
Q.push(v);
}
}
}
return deep[T];
}
ll dfs(int u, ll dist)
{
if(u == T) return dist;
ll f, di;
for(int &i=cur[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to; f = edge[i].flow;
if(f && deep[v] == deep[u] + 1)
{
di = dfs(v, min(dist, f));
if(di)
{
edge[i].flow -= di; edge[i ^ 1].flow += di;
return di;
}
}
}
return 0;
}
inline ll Dinic()
{
ll ans = 0, tmp;
while(bfs())
{
for(int i=0; i<=node; i++) cur[i] = head[i];
while((tmp = dfs(S, INF))) ans += tmp;
}
return ans;
}
} mf;
struct TP_Sort
{
bool can_use[maxN]; //can_use == can use this point
int du[maxN], node;
vector<int> vt[maxN];
inline void _add_E(int u, int v) { vt[u].push_back(v); }
queue<int> Q;
void solve()
{
for(int i=1, len, v; i<=node; i++)
{
len = (int)vt[i].size();
for(int j=0; j<len; j++)
{
v = vt[i][j];
du[v]++;
}
}
for(int i=1; i<=node; i++) if(!du[i]) { Q.push(i); can_use[i] = true; }
while(!Q.empty())
{
int u = Q.front(); Q.pop();
int len = (int)vt[u].size();
for(int i=0, v; i<len; i++)
{
v = vt[u][i]; du[v]--;
if(!du[v])
{
can_use[v] = true;
Q.push(v);
}
}
}
}
} tj;
inline void init()
{
cnt = 0; mf.S = 0; mf.T = N * M + 1; mf.node = N * M + 2; tj.node = N * M;
for(int i=0; i<=mf.node; i++) head[i] = -1;
}
int main()
{
scanf("%d%d", &N, &M);
init();
for(int i=0, id, nex_cnt, x, y, nd; i<N; i++)
{
for(int j=1; j<=M; j++)
{
id = i * M + j;
if(j ^ 1) tj._add_E(id, id - 1);
scanf("%d%d", &score[id], &nex_cnt);
while(nex_cnt--)
{
scanf("%d%d", &x, &y); nd = x * M + y + 1;
if(nd == id) { zihuan[id] = true; continue; }
tj._add_E(id, nd);
}
}
}
tj.solve();
ll sum = 0;
for(int i=0, id; i<N; i++)
{
for(int j=1; j<=M; j++)
{
id = i * M + j;
if(!tj.can_use[id]) continue;
if(score[id] >= 0) { _add(mf.S, id, score[id]); sum += score[id]; }
else _add(id, mf.T, -score[id]);
int len = (int)tj.vt[id].size();
for(int k=0, v; k<len; k++)
{
v = tj.vt[id][k];
if(!tj.can_use[v]) continue;
_add(v, id, INF);
}
}
}
printf("%lld\n", sum - mf.Dinic());
return 0;
}
/*
3 2
10 0
20 0
-10 0
-5 1 0 0
100 1 2 1
100 0
*/