hdu - 4975 - A simple Gaussian elimination problem.（最大流）

题意：给一个N行M列的数字矩阵的行和以及列和，每个元素的大小不超过9，问这样的矩阵是否存在，是否唯一N(1 ≤ N ≤ 500) , M(1 ≤ M ≤ 500)。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4975

——>>方法如：http://blog.csdn.net/scnu_jiechao/article/details/40658221

先做hdu - 4888，再来做此题的时候，感觉这题好 SB 呀，将代码提交后自己就 SB 了，咋 TLE 了？？


此题卡时间比较严。。可以在判环的地方优化一下4888的代码，因为原选建好的图中的行到列的边，在判环的时候，每条边会被判 (N - 1) * (M - 1) + 1 次，如果先对残量网络重新建图，那么就只会在建图的时候被判一次，差距甚远！

注：输入开挂会RE，猜想输入数据中有负数。。


#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using std::min;
using std::queue;

const int MAXN = 500 * 2 + 10;
const int MAXM = 500 * 500 + 2 * MAXN;
const int INF = 0x3f3f3f3f;

struct EDGE
{
int from;
int to;
int cap;
int flow;
int nxt;
} edge[MAXM << 1];

int N, M, kase;
int sum;
int S, T;
int hed[MAXN], ecnt;
int cur[MAXN], h[MAXN];
bool impossible, bUnique;

void Init()
{
impossible = false;
bUnique = true;
ecnt = 0;
memset(hed, -1, sizeof(hed));
}

void AddEdge(int u, int v, int cap)
{
edge[ecnt].from = u;
edge[ecnt].to = v;
edge[ecnt].cap = cap;
edge[ecnt].flow = 0;
edge[ecnt].nxt = hed[u];
hed[u] = ecnt++;
edge[ecnt].from = v;
edge[ecnt].to = u;
edge[ecnt].cap = 0;
edge[ecnt].flow = 0;
edge[ecnt].nxt = hed[v];
hed[v] = ecnt++;
}

bool Bfs()
{
memset(h, -1, sizeof(h));
queue<int> qu;
qu.push(S);
h[S] = 0;
while (!qu.empty())
{
int u = qu.front();
qu.pop();
for (int e = hed[u]; e != -1; e = edge[e].nxt)
{
int v = edge[e].to;
if (h[v] == -1 && edge[e].cap > edge[e].flow)
{
h[v] = h[u] + 1;
qu.push(v);
}
}
}

return h[T] != -1;
}

int Dfs(int u, int cap)
{
if (u == T || cap == 0) return cap;

int flow = 0, subFlow;
for (int e = cur[u]; e != -1; e = edge[e].nxt)
{
cur[u] = e;
int v = edge[e].to;
if (h[v] == h[u] + 1 && (subFlow = Dfs(v, min(cap, edge[e].cap - edge[e].flow))) > 0)
{
flow += subFlow;
edge[e].flow += subFlow;
edge[e ^ 1].flow -= subFlow;
cap -= subFlow;
if (cap == 0) break;
}
}

return flow;
}

int Dinic()
{
int maxFlow = 0;

while (Bfs())
{
memcpy(cur, hed, sizeof(hed));
maxFlow += Dfs(S, INF);
}

return maxFlow;
}

int ReadInt()
{
int ret = 0;
char ch;

while ((ch = getchar()) && ch >= '0' && ch <= '9')
{
ret = ret * 10 + ch - '0';
}

return ret;
}

void Read()
{
int r, c;
int rsum = 0, csum = 0;

scanf("%d%d", &N, &M);
S = 0;
T = N + M + 1;
getchar();
for (int i = 1; i <= N; ++i)
{
// r = ReadInt();
scanf("%d", &r);
rsum += r;
AddEdge(S, i, r);
}
for (int i = 1; i <= M; ++i)
{
// c = ReadInt();
scanf("%d", &c);
csum += c;
AddEdge(i + N, T, c);
}

if (rsum != csum)
{
impossible = true;
return;
}

sum = rsum;
for (int i = 1; i <= N; ++i)
{
for (int j = M; j >= 1; --j)
{
AddEdge(i, j + N, 9);
}
}
}

void CheckPossible()
{
if (impossible) return;
if (Dinic() != sum)
{
impossible = true;
}
}

void AddEdge(int u, int v)
{
edge[ecnt].to = v;
edge[ecnt].nxt = hed[u];
hed[u] = ecnt++;
}

void ReBuild()
{
memset(hed, -1, sizeof(hed));
int total = ecnt;
ecnt = 0;
for (int e = 0; e < total; ++e)
{
if (edge[e].cap > edge[e].flow)
{
AddEdge(edge[e].from, edge[e].to);
}
}
}

bool vis[MAXN];
bool CheckCircle(int x, int f)
{
vis[x] = true;
for (int e = hed[x]; e != -1; e = edge[e].nxt)
{
int v = edge[e].to;
if (v != f)
{
if (vis[v]) return true;
if (CheckCircle(v, x)) return true;
}
}
vis[x] = false;
return false;
}

void CheckUnique()
{
if (impossible) return;
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= N; ++i)
{
if (CheckCircle(i, -1))
{
bUnique = false;
return;
}
}
}

void Output()
{
printf("Case #%d: ", ++kase);
if (impossible)
{
puts("So naive!");
}
else if (bUnique)
{
puts("So simple!");
}
else
{
puts("So young!");
}
}

int main()
{
int T;

scanf("%d", &T);
while (T--)
{
Init();
Read();
CheckPossible();
ReBuild();
CheckUnique();
Output();
}

return 0;
}