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#include <cstdio> | |
#include <cstring> | |
#include <queue> | |
#include <vector> | |
using namespace std; | |
#define INF 9999999 | |
#define MAX 2*52*52+10 | |
vector<int> edge[MAX]; | |
int cap[MAX][MAX], flow[MAX][MAX]; | |
int bottleneck[MAX], pre[MAX]; | |
void BuildGraph(int S, int T, int H, int W); | |
int MaxFlow(int S, int T, int H, int W); | |
int main() | |
{ | |
int Case, H, W, B, x, y; | |
scanf("%d", &Case); | |
while (Case--) { | |
scanf("%d %d %d", &H, &W, &B); | |
int S = 0, // super source | |
T = 1; // super sink | |
for (int i = 0; i <= 2*(H+1)*W; ++i) edge[i].clear(); | |
memset(cap, 0, sizeof(cap)); | |
memset(flow, 0, sizeof(flow)); | |
BuildGraph(S, T, H, W); | |
for (int i = 0; i < B; ++i) { | |
scanf("%d %d", &x, &y); | |
cap[S][(x*H+y)*2] = 1; | |
edge[S].push_back((x*H+y)*2); | |
} | |
if (MaxFlow(S, T, H, W) == B) puts("possible"); | |
else puts("not possible"); | |
} | |
} | |
void BuildGraph(int S, int T, int H, int W) | |
{ | |
for (int i = 1; i <= H; ++i) { | |
for (int j = 1; j <= W; ++j) { | |
cap[(i*H+j)*2][(i*H+j)*2+1] = 1; // u to u' | |
edge[(i*H+j)*2].push_back((i*H+j)*2+1); | |
cap[(i*H+j)*2+1][((i-1)*H+j)*2] = 1; // u' to up | |
edge[(i*H+j)*2+1].push_back(((i-1)*H+j)*2); | |
cap[(i*H+j)*2+1][((i+1)*H+j)*2] = 1; // u' to down | |
edge[(i*H+j)*2+1].push_back(((i+1)*H+j)*2); | |
cap[(i*H+j)*2+1][(i*H+j-1)*2] = 1; // u' to left | |
edge[(i*H+j)*2+1].push_back((i*H+j-1)*2); | |
cap[(i*H+j)*2+1][(i*H+j+1)*2] = 1; // u' to right | |
edge[(i*H+j)*2+1].push_back((i*H+j+1)*2); | |
if (i == 1 || j == 1 || i == H || j == W) {// if on the edge | |
cap[(i*H+j)*2+1][T] = 1; // connect u' to the T | |
edge[(i*H+j)*2+1].push_back(T); | |
} | |
} | |
} | |
} | |
int MaxFlow(int S, int T, int H, int W) | |
{ | |
int result = 0; | |
while (1) { | |
memset(bottleneck, 0, sizeof(bottleneck)); | |
bottleneck[S] = INF; | |
queue<int> Q; | |
Q.push(S); | |
// bfs | |
while (!Q.empty() && !bottleneck[T]) { | |
int now = Q.front(); | |
Q.pop(); | |
for (int nxt : edge[now]) { | |
if (!bottleneck[nxt] && cap[now][nxt] > flow[now][nxt]) { | |
Q.push(nxt); | |
pre[nxt] = now; | |
bottleneck[nxt] = min(bottleneck[now], cap[now][nxt] - flow[now][nxt]); | |
} | |
} | |
} | |
if (bottleneck[T] == 0) break; | |
for (int now = T; now != S; now = pre[now]) { | |
flow[pre[now]][now] += bottleneck[T]; | |
flow[now][pre[now]] -= bottleneck[T]; | |
} | |
result += bottleneck[T]; | |
} | |
return result; | |
} |
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