HDU 4085 Peach Blossom Spring

HDU_4085

去年去北京时用网络流死磕这题导致了全场悲剧，今年再做依旧毫无头绪，找到题解一看瞬间震惊了，斯坦纳树(Stenier tree)是什么东东（不过后来发现其实解法就是个状态压缩dp，只不过要借助spfa完成状态转移）……于是只好一点点学习别人的代码了，现在自己理解的还不是很透彻，推荐一篇感觉写得还不错的文章：http://endlesscount.blog.163.com/blog/static/821197872012525113427573/，回头再多做些有关的题目加深一下理解。

#include<stdio.h>
#include<string.h>
#define MAXD 60
#define MAXM 2010
#define MAXQ 2000010
#define ST 1034
#define INF 0x3f3f3f3f
const int Q = 2000000;
int N, M, K, bit[MAXD], first[MAXD], e, next[MAXM], v[MAXM], w[MAXM];
int q[MAXQ], inq[MAXD][ST], front, rear, f[MAXD][ST], dp[ST];
void add(int x, int y, int z)
{
v[e] = y, w[e] = z;
next[e] = first[x], first[x] = e ++;
}
void init()
{
int i, x, y, z;
scanf("%d%d%d", &N, &M, &K);
memset(first, -1, sizeof(first));
e = 0;
for(i = 0; i < M; i ++)
{
scanf("%d%d%d", &x, &y, &z);
add(x, y, z), add(y, x, z);
}
}
int Min(int x, int y)
{
return x < y ? x : y;
}
void spfa()
{
int i, x, st, y, nst;
while(front != rear)
{
x = q[front] & 1023, st = q[front] >> 10;
inq[x][st] = 0;
++ front > Q ? front = 0 : 0;
for(i = first[x]; i != -1; i = next[i])
{
y = v[i], nst = st | bit[y];
if(f[x][st] + w[i] < f[y][nst])
{
f[y][nst] = f[x][st] + w[i];
if(nst == st && !inq[y][nst])
{
q[rear ++] = nst << 10 | y, inq[y][nst] = 1;
rear > Q ? 0 : 0;
}
}
}
}
}
int check(int st)
{
int i, a = 0;
for(i = 0; i < K; i ++)
{
if(st & 1 << i) ++ a;
if(st & 1 << K + i) -- a;
}
return a == 0;
}
void solve()
{
int i, j, k, nn = 1 << 2 * K;
memset(f, 0x3f, sizeof(f));
memset(bit, 0, sizeof(bit));
for(i = 1; i <= K; i ++)
{
bit[i] = 1 << (i - 1), f[i][bit[i]] = 0;
bit[N - K + i] = 1 << (K + i - 1), f[N - K + i][bit[N - K + i]] = 0;
}
front = rear = 0;
memset(inq, 0, sizeof(inq));
for(i = 0; i < nn; i ++)
{
for(j = 1; j <= N; j ++)
{
for(k = i - 1 & i; k; k = k - 1 & i) // 枚举i的所有子集
f[j][i] = Min(f[j][i], f[j][k | bit[j]] + f[j][i - k | bit[j]]);
if(f[j][i] < INF)
{
q[rear ++] = i << 10 | j, inq[j][i] = 1;
rear > Q ? rear = 0 : 0;
}
}
spfa();
}
memset(dp, 0x3f, sizeof(dp));
for(i = 0; i < nn; i ++)
for(j = 1; j <= N; j ++)
dp[i] = Min(dp[i], f[j][i]);
for(i = 0; i < nn; i ++)
if(check(i))
{
for(j = i - 1 & i; j; j = j - 1 & i)
if(check(j))
dp[i] = Min(dp[i], dp[j] + dp[i - j]);
}
if(dp[nn - 1] == INF)
printf("No solution\n");
else
printf("%d\n", dp[nn - 1]);
}
int main()
{
int t;
scanf("%d", &t);
while(t --)
{
init();
solve();
}
return 0;
}