题目链接 P2055 [ZJOI2009]假期的宿舍
关于这道题的做法,我所使用的算法可能会复杂了,实际上我们要先理清楚题意(尽管它是道中文题)。
首先,认识关系是没有传递性的,譬如A认识B,B认识C,C认识D,C和D要回家了,A和B是外校的,不代表B能先睡C的给A暖暖炕头再搬去C认识的D的床位,然后A睡B之前睡的C的床位。
若A认识B,那么只能A去睡B的床,不是说B认识X,A就能去睡X的床位,认识没有传递性。
那么,我们可以认为是人去匹配床位,当然,本身这所学校的应当也可以睡在自己原本的床位,于是就变成了拆点分成人和床的这么一个二分图的最大匹配问题了,若最大匹配数等同于需要在学校里的人数(不回家和外校的人的总和),那么就是可行解。
#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 0x3f3f3f3f
#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
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e2 + 7, maxM = 2e4 + 7;
int N, school[maxN], live[maxN];
int S, T, head[maxN], cnt, cur[maxN];
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 gap[maxN], d[maxN], que[maxN], ql, qr, node;
inline void init()
{
for(int i=0; i<=node + 1; i++)
{
gap[i] = d[i] = 0;
cur[i] = head[i];
}
++gap[d[T] = 1];
que[ql = qr = 1] = T;
while(ql <= qr)
{
int x = que[ql ++];
for(int i=head[x], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(!d[v]) { ++gap[d[v] = d[x] + 1]; que[++qr] = v; }
}
}
}
inline ll aug(int x, ll FLOW)
{
if(x == T) return FLOW;
int flow = 0;
for(int &i=cur[x], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(d[x] == d[v] + 1)
{
ll tmp = aug(v, min(FLOW, edge[i].flow));
flow += tmp; FLOW -= tmp; edge[i].flow -= tmp; edge[i ^ 1].flow += tmp;
if(!FLOW) return flow;
}
}
if(!(--gap[d[x]])) d[S] = node + 1;
++gap[++d[x]]; cur[x] = head[x];
return flow;
}
inline ll max_flow()
{
init();
ll ret = aug(S, INF);
while(d[S] <= node) ret += aug(S, INF);
return ret;
}
} mf;
inline void init()
{
S = 0; T = N + N + 1; mf.node = T + 1;
cnt = 0;
for(int i=0; i<=mf.node; i++) head[i] = -1;
}
int main()
{
int Cas; scanf("%d", &Cas);
while(Cas--)
{
scanf("%d", &N);
init();
int need = 0;
for(int i=1; i<=N; i++)
{
scanf("%d", &school[i]);
if(!school[i]) { _add(S, i, 1); need++; }
else _add(i, N + i, 1);
}
for(int i=1; i<=N; i++)
{
scanf("%d", &live[i]);
if(school[i])
{
if(!live[i]) { _add(S, i, 1); need++; }
_add(N + i, T, 1);
}
}
for(int i=1, friend_ship; i<=N; i++)
{
for(int j=1; j<=N; j++)
{
scanf("%d", &friend_ship);
if(friend_ship) _add(i, N + j, 1);
}
}
ll max_Flow = mf.max_flow();
if(max_Flow == need) printf("^_^\n");
else printf("T_T\n");
}
return 0;
}