传送门：https://www.luogu.org/problemnew/show/P2805
题意
有一个n * m的地图，你可以操纵僵尸从地图的右边向左边走，走的一些地方是有能量值的，有些地方会被一些植物保护起来不能走，只有先吃掉特定植物才能走一些地方。求最大可能拿到的能量值和
思路
最大权闭合子图，由于僵尸只能从一行的右边一步一步走到左边，所以每个格子向右边连一条inf的边（表示选了这个点，右边这个点必选），然后有保护的原因，从被保护的格子向保护的格子连一条inf的边。然后就是最大权闭合子图的操作，源点向正权值格子连容量为这个权值的边，负权值的格子向终点连容量为这个权值绝对值的边。
由于图中有环的存在，我们要把环上以及环之后的点都抹去。
然后跑一遍dinic（），算出最小割，用总的正权值 - 这个最小割就是答案。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> /* ⊂_ヽ ＼＼ Λ＿Λ 来了老弟 ＼('ㅅ') > ⌒ヽ / へ＼ / / ＼＼ ﾚ ノ ヽ_つ / / / /| ( (ヽ | |、＼ | 丿 ＼ ⌒) | | ) / 'ノ ) Lﾉ */ using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define boost ios::sync_with_stdio(false);cin.tie(0) #define rep(a, b, c) for(int a = (b); a <= (c); ++ a) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } inline void cmax(int &x,int y){if(x<y)x=y;} inline void cmax(ll &x,ll y){if(x<y)x=y;} inline void cmin(int &x,int y){if(x>y)x=y;} inline void cmin(ll &x,ll y){if(x>y)x=y;} /*-----------------------showtime----------------------*/ const int maxn = 609; int val[maxn],vis[maxn],du[maxn]; vector<int>mp[maxn]; vector<pii>ptt[maxn];//受保护的点。 int n,m; void topo(){ queue<int>que; for(int i=1; i<=n*m; i++){ if(du[i] == 0) que.push(i); } while(!que.empty()){ int u = que.front(); que.pop(); vis[u] = 1; for(int i=0; i<mp[u].size(); i++){ int v = mp[u][i]; du[v] --; if(du[v] == 0) que.push(v); } } } struct E{ int v,w; int nxt; }edge[500009]; int gtot = 0, head[maxn]; void addedge(int u,int v,int w){ edge[gtot].v = v; edge[gtot].w = w; edge[gtot].nxt = head[u]; head[u] = gtot++; edge[gtot].v = u; edge[gtot].w = 0; edge[gtot].nxt = head[v]; head[v] = gtot++; } int dis[maxn],cur[maxn]; bool bfs(int s,int t){ memset(dis, inf, sizeof(dis)); dis[s] = 0; queue<int>que; que.push(s); for(int i=s; i<=t; i++) cur[i] = head[i]; while(!que.empty()){ int u = que.front(); que.pop(); for(int i=head[u]; ~i; i = edge[i].nxt){ int v = edge[i].v, w = edge[i].w; if(w > 0 && dis[v] > dis[u] + 1){ dis[v] = dis[u] + 1; que.push(v); } } } return dis[t] < inf; } int dfs(int u,int t,int maxflow){ if(u == t || maxflow == 0) return maxflow; for(int i=cur[u]; ~i; i = edge[i].nxt){ cur[u] = i; int v = edge[i].v, w = edge[i].w; if(w > 0 && dis[v] == dis[u] + 1) { int f = dfs(v, t, min(w, maxflow)); if(f > 0) { edge[i].w -= f; edge[i^1].w += f; return f; } } } return 0; } int dinic(int s,int t){ int flow = 0; while(bfs(s, t)){ while(int f = dfs(s,t,inf)) flow += f; } return flow; } int main(){ memset(head, -1, sizeof(head)); scanf("%d%d", &n, &m); for(int i=1; i<=n; i++){ for(int j=1; j<=m; j++){ int u = (i-1)*m + j, v = u + 1; scanf("%d", &val[u]); if(j < m){ du[u]++;mp[v].pb(u);//v -> u } int q; scanf("%d", &q); while(q -- ){ int x,y; scanf("%d%d", &x, &y); x++,y++; int p = (x-1)*m+y; du[p]++; mp[u].pb(p); ptt[u].pb(pii(x,y)); } } } int s = 0, t = n*m+1; topo(); int sum = 0; for(int i=1; i<=n; i++){ for(int j=1; j<=m; j++){ int u = (i-1)*m + j, v = u + 1; if(vis[u] == 0) continue; if(j < m && vis[v])addedge(u, v, inf); if(val[u] >= 0) addedge(s, u, val[u]), sum += val[u]; else addedge(u, t, -1*val[u]); for(int k=0; k < ptt[u].size(); k++){ int x = ptt[u][k].fi, y = ptt[u][k].se; int p = (x - 1) * m + y; addedge(p, u, inf); } } } printf("%d\n", sum - dinic(s,t)); return 0; }