BZOJ1458 士兵占领 【带上下界网络流】

题目链接

BZOJ1458

题解

对行列分别建边，拆点，设置流量下限

然后\(S\)向行连边\(inf\)，列向\(T\)连边\(inf\)，行列之间如果没有障碍，就连边\(1\)

然后跑最小可行流即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 505,maxm = 500005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int h[maxn],ne = 1;
struct EDGE{int to,nxt,f;}ed[maxm];
inline void build(int u,int v,int f){
	ed[++ne] = (EDGE){v,h[u],f}; h[u] = ne;
	ed[++ne] = (EDGE){u,h[v],0}; h[v] = ne;
}
int vis[maxn],used[maxn],cur[maxn],d[maxn],S,T,now;
int q[maxn],head,tail;
bool bfs(){
	vis[S] = now; d[S] = 0; q[head = tail = 0] = S;
	int u;
	while (head <= tail){
		u = q[head++];
		Redge(u) if (ed[k].f && vis[to = ed[k].to] != now){
			d[to] = d[u] + 1; vis[to] = now;
			if (to == T) return true;
			q[++tail] = to;
		}
	}
	return vis[T] == now;
}
int dfs(int u,int minf){
	if (u == T || !minf) return minf;
	int flow = 0,f,to;
	if (used[u] != now) cur[u] = h[u],used[u] = now;
	for (int& k = cur[u]; k; k = ed[k].nxt)
		if (vis[to = ed[k].to] == now && d[to] == d[u] + 1 && (f = dfs(to,min(ed[k].f,minf)))){
			ed[k].f -= f; ed[k ^ 1].f += f;
			flow += f; minf -= f;
			if (!minf) break;
		}
	return flow;
}
int maxflow(){
	int flow = 0; now = 1;
	while (bfs()){
		flow += dfs(S,INF);
		now++;
	}
	return flow;
}
int n,m,K,L[105],C[105],del[105],de[105],g[105][105];
int main(){
	n = read(); m = read(); K = read();
	REP(i,n) L[i] = read(),del[i] = m;
	REP(i,m) C[i] = read(),de[i] = n;
	int a,b;
	while (K--){
		a = read(); b = read();
		g[a][b] = true;
		del[a]--;
		de[b]--;
	}
	S = 0; T = ((n + m) << 1) + 3;
	int SS = (n + m) << 1 | 1,TT = SS + 1,E = n + m;
	REP(i,n){
		build(SS,i,INF);
		build(S,i + E,L[i]);
		build(i,T,L[i]);
		if (del[i] > L[i]) build(i,i + E,del[i] - L[i]);
		REP(j,m) if (!g[i][j]){
			build(i + E,n + j,1);
		}
	}
	REP(i,m){
		build(n + i + E,TT,INF);
		build(S,n + i + E,C[i]);
		build(n + i,T,C[i]);
		if (de[i] > C[i]) build(n + i,n + i + E,de[i] - C[i]);
	}
	maxflow();
	build(TT,SS,INF);
	maxflow();
	Redge(S) if (ed[k].f){puts("JIONG!"); return 0;}
	Redge(T) if (ed[k ^ 1].f){puts("JIONG!"); return 0;}
	printf("%d\n",ed[h[TT] ^ 1].f);
	return 0;
}