HDU 3605 - Escape (多重匹配 | 最大流 & Dinic)
思路
折腾一天了这题。
10W的数据量,普通的网络流建图会TLE。
学习了缩点。就是把重复的点压缩,然后容量上增加。一共只有 (1 « 10)种状况。
后来用多重匹配写的时候竟然无限RE,现在还没找出原因。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 1500 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
struct EDGE
{
int from, to, cap, flow;
};
struct DINIC
{
int d[MAXN], cur[MAXN], vis[MAXN], st, ed;
vector<EDGE> edges;
vector<int> G[MAXN];
void add_edge(int from, int to, int cap)
{
EDGE a = {from, to, cap, 0};
EDGE b = {to, from, 0, 0};
edges.PB(a);
edges.PB(b);
int m = SZ(edges);
G[from].PB(m - 2); G[to].PB(m - 1);
}
void init(int st, int ed)
{
this->st = st, this->ed = ed;
edges.clear();
for(int i = 0; i <= ed; i++) G[i].clear();
}
int maxFlow()
{
int flow = 0;
while (BFS())
{
MS(cur, 0);
flow += DFS(st, INF);
}
return flow;
}
bool BFS()
{
MS(vis, 0);
queue<int> qu;
qu.push(st);
d[st] = 0; vis[st] = 1;
while (!qu.empty())
{
int x = qu.front(); qu.pop();
for (int i = 0; i < SZ(G[x]); i++)
{
EDGE &e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow)
{
vis[e.to] = 1;
d[e.to] = d[x] + 1;
qu.push(e.to);
}
}
}
return vis[ed];
}
int DFS(int x, int a)
{
if (x == ed || a == 0) return a;
int flow = 0, f;
for (int &i = cur[x]; i < SZ(G[x]); 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;
}
}maxFlow;
int state[1500];
int main()
{
//ROP;
int npeo, m, i, j;
while (~scanf("%d%d", &npeo, &m))
{
MS(state, 0);
int ed = (1 << m) + m, st = 0;
maxFlow.init(st, ed);
for (i = 1; i <= npeo; i++)
{
int sum = 0, tmp;
for (j = 0; j < m; j++)
{
scanf("%d", &tmp);
if (tmp) sum += (1 << j);
}
state[sum]++;
}
int tmp;
for (i = (1 << m); i < (1 << m) + m; i++)
{
scanf("%d", &tmp);
maxFlow.add_edge(i, ed, tmp);
}
for (i = 1; i < (1 << m); i++)
{
if (state[i])
{
maxFlow.add_edge(st, i, state[i]);
for (int j = 0; j < m; j++)
if (i & (1 << j)) maxFlow.add_edge(i, (1 << m) + j, state[i]);
}
}
int flow = 0;
printf("%s\n", (maxFlow.maxFlow() == npeo ? "YES" : "NO"));
}
return 0;
}