14更多学校的第二个问题

网络流量   分别以行,列作为结点建图

i行表示的结点到j列表示的结点的流量便是(i, j)的值

跑遍最大流   若满流了便是有解   推断是否unique  就是在残余网络中dfs。走能够添加流量的边,找到环即不唯一

dfs的时候一定要回溯!!

。。

#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
#include <bitset>
#include <fstream>
using namespace std; //LOOP
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++) //OTHER
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define all(x) (x).begin(),(x).end() //INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RS(s) scanf("%s", s) //OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(n) printf("%s\n", n) //debug
//#define online_judge
#ifndef online_judge
#define dt(a) << (#a) << "=" << a << " "
#define debugI(a) cout dt(a) << endl
#define debugII(a, b) cout dt(a) dt(b) << endl
#define debugIII(a, b, c) cout dt(a) dt(b) dt(c) << endl
#define debugIV(a, b, c, d) cout dt(a) dt(b) dt(c) dt(d) << endl
#define debugV(a, b, c, d, e) cout dt(a) dt(b) dt(c) dt(d) dt(e) << endl
#else
#define debugI(v)
#define debugII(a, b)
#define debugIII(a, b, c)
#define debugIV(a, b, c, d)
#endif #define sqr(x) (x) * (x)
typedef long long LL;
typedef unsigned long long ULL;
typedef vector <int> VI;
const double eps = 1e-9;
const int MOD = 1000000007;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int maxn = 900; bool use[maxn];
struct Edge{
int from, to, cap, flow;
};
int MAX; struct Dinic{
int n, m ,s, t;
vector<Edge> edges;
VI G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn] ; void init(int nn)
{
this->n = nn;
REP(i, n) G[i].clear();
edges.clear();
} void addEdge(int from, int to, int cap)
{
edges.PB((Edge){from, to, cap, 0});
edges.PB((Edge){to, from, 0, 0});
m = edges.size();
G[from].PB(m - 2);
G[to].PB(m - 1);
} bool bfs()
{
CLR(vis, 0);
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = 1;
while (!Q.empty())
{
int x = Q.front();
Q.pop();
REP(i, G[x].size())
{
Edge& e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow)
{
vis[e.to] = 1;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
} int dfs(int x, int a)
{
if (x == t || a == 0) return a;
int flow = 0, f;
for (int& i = cur[x]; i < G[x].size(); i++)
{
Edge& e = edges[G[x][i]];
if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)
{
e.flow += f;
edges[G[x][i] ^ 1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
} int maxflow(int s, int t)
{
this-> s = s, this-> t = t;
int flow = 0;
while (bfs())
{
CLR(cur, 0);
flow += dfs(s, INF);
}
return flow;
} bool visit(int u, int fa)
{
if (u == 0 || u == MAX) return false;
use[u] = 1;
REP(i, G[u].size())
{
Edge& e = edges[G[u][i]];
// debugII(e.to, use[e.to]);
if (e.to != fa && e.cap > e.flow)
if (use[e.to] || visit(e.to, u))
return true;
}
use[u] = 0;
return false;
} }di; int main()
{
int n, m, k;
while (~RIII(n, m, k))
{
int x, sum1 = 0, sum2 = 0;
MAX = n + m + 1;
di.init(n + m + 2);
FE(i, 1, n)
{
RI(x);
sum1 += x;
di.addEdge(0, i, x);
}
FE(i, n + 1, n + m)
{
RI(x);
sum2 += x;
di.addEdge(i, n + m + 1, x);
}
FE(i, 1, n)
FE(j, n + 1, n + m)
di.addEdge(i, j, k);
if (sum2 != sum1)
{
puts("Impossible");
continue;
}
int ans = di.maxflow(0, n + m + 1);
if (ans < sum1)
{
puts("Impossible");
continue;
}
FE(i, 1, n)
{
CLR(use, 0);
if (di.visit(i, -1))
{
puts("Not Unique");
goto end;
}
}
puts("Unique");
for (int i = 2 * n + 2 * m; i < di.edges.size(); i += 2)
{
printf("%d", di.edges[i].flow);
if (i + 2 >= di.edges.size())
printf("\n");
else
printf(" ");
}
end:;
}
return 0;
}

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