C - Dynamic Graph Matching HDU - 6321
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状压dp,先记录偶数状态,然后直接暴力转移即可。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#pragma GCC optimize(2)
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
int num = 0;
int T, n, m;
int sta[5000];
int edges[5000];
ll dp[10][2000];
const int mod = 1e9 + 7;
void init()
{
up(i, 1, (1 << 10))
{
int cnt = 0;
dwd(j, 10, 0)
{
if (i >> j & 1)cnt++;
}
if (cnt % 2 == 0)
{
sta[num] = i;
edges[num++] = cnt / 2;
}
}
}
ll ans[30004];
int main()
{
T = read();
init();
while (T--)
{
int cur = 0;
n = read(), m = read();
upd(i, 0, (1 << n))dp[cur][i] = dp[cur ^ 1][i] = 0;
dp[cur][0] = dp[cur ^ 1][0] = 1;
int status = (1 << n) - 1;
int u, v;
char op[2];
upd(p, 1, m)
{
scanf("%s %d %d", op, &u, &v); u--, v--;
if (op[0] == '+')
{
cur = cur ^ 1;
int temp = ((1 << u) | (1 << v));
up(i, 0, num)
{
int num_s= sta[i];
if (num_s > status)break;
dp[cur][num_s] = dp[cur ^ 1][num_s];
if ((temp&num_s)==temp)
{
dp[cur][num_s] = (dp[cur][num_s] + dp[cur ^ 1][num_s^temp]) % mod;
}
ans[edges[i]] = (ans[edges[i]] + dp[cur][num_s]) % mod;
}
upd(i, 1, n / 2)printf("%lld%s", ans[i],i==n/2?"
":" ");
}
else{
cur = cur ^ 1;
int temp = ((1 << u) | (1 << v));
up(i, 0, num)
{
int num_s = sta[i];
if (num_s > status)break;
dp[cur][num_s] = dp[cur ^ 1][num_s];
if ((temp&num_s)==temp)
{
dp[cur][num_s] = (dp[cur][num_s] - dp[cur ^ 1][num_s^temp] + mod) % mod;
}
ans[edges[i]] = (ans[edges[i]] + dp[cur][num_s]) % mod;
}
upd(i, 1, n / 2)printf("%lld%s", ans[i], i == n / 2 ? "
" : " ");
}
upd(i, 0, 10)ans[i] = 0;
}
}
return 0;
}
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