[BZOJ 3691] 游行

http://www.lydsy.com/JudgeOnline/problem.php?id=3691

Solution :

被这题折磨了好多天了。。。问了问学长。。。惊讶的发现wmd写过题解。。。。。。我。。。。。

首先我们跑一发floyd。。。算出任意两点最短路。。。于是我们可以假设路径不相交。。。然后。。。往最小点覆盖方向想一想。。。发现一个匹配模型。。。我每次询问都想把C加入图里强行算一遍。。。然后就做不下去了。。。

正解需要灵光一现一下。。。实际上根本没有必要把C加入图里。。。有几个点没有被覆盖。。我们就需要支付几个C。。。于是我们初始化的时候跑一次费用流，记录下每增广一点流量的时候的cost。。。必然单调。。所以对于每次讯问。。。我们在cost数组里二分出一个位置。。cost之前的增广。。。cost之后的支付C。。。容易发现这是最优的。。。

想通这些就去码了。。。数组开大 1 A 了。。。爽啊。。。

#include <cstdio>
#include <cstring>
#include <cstdlib>

#define rep(i, x, y) for (int i = (x), _ = (y); i <= _; ++i)
#define down(i, x, y) for (int i = (x), _ = (y); i >= _; --i)
#define x first
#define y second

using namespace std;
typedef long long LL;

template<typename T> inline void upmax(T & x, T y) { x < y ? x = y : 0; }
template<typename T> inline void upmin(T & x, T y) { x > y ? x = y : 0; }

template<typename T>
inline void read(T & x)
{
char c;
while ((c = getchar()) < '0' || c > '9') ;
for (x = c - '0'; (c = getchar()) >= '0' && c <= '9'; x = x * 10 - '0' + c) ;
}

const int inf = 0x3f3f3f3f;
const int N = 511;
const int M = 1e5 + 10;

struct edge
{
int to, c, w, n;
} e[M * 2];

int head
, tot = 1;

inline void addEdge(int x, int y, int c, int w)
{
e[++tot] = (edge) {y, c, w, head[x]}, head[x] = tot;
e[++tot] = (edge) {x, 0, -w, head[y]}, head[y] = tot;
}

int pa
, d
, S, T;

bool SPFA()
{
static bool inq
;
static int que[N + 5], fir, las;

fir = las = 0;
memset(d, 0x3f, sizeof(int) * (T + 5));

d[S] = 0;
que[++las] = S;

while (fir != las)
{
int u = que[++fir];
fir == N ? fir = 0 : 0;
inq[u] = 0;
for (int p = head[u]; p; p = e[p].n) if (e[p].c)
{
int to = e[p].to;
if (d[to] > d[u] + e[p].w)
{
pa[to] = p;
d[to] = d[u] + e[p].w;
if (!inq[to])
{
inq[to] = 1;
que[++las] = to;
las == N ? las = 0 : 0;
}
}
}
}

return d[T] < inf;
}

int augment()
{
int x = T, a = inf;
while (x != S)
{
upmin(a, e[pa[x]].c);
x = e[pa[x] ^ 1].to;
}
x = T;
while (x != S)
{
e[pa[x]].c -= a;
e[pa[x] ^ 1].c += a;
x = e[pa[x] ^ 1].to;
}
return d[T] * a;
}

int f

;
int cost
, sum
, cnt;

int main()
{
#ifdef LX_JUDGE
freopen("in.txt", "r", stdin);
#endif

int n, m, Q;
read(n), read(m), read(Q);

S = 0, T = n + n + 1;

memset(f, 0x3f, sizeof(f));

int x, y, w;

rep (i, 1, m)
{
read(x), read(y), read(w);
upmin(f[x][y], w);
}

rep (k, 1, n) rep (i, 1, n) rep (j, 1, n)
upmin(f[i][j], f[i][k] + f[k][j]);

rep (i, 1, n)
{
addEdge(S, i, 1, 0);
addEdge(i + n, T, 1, 0);
}

rep (i, 1, n) rep (j, 1, n) if (f[i][j] < inf)
addEdge(i, j + n, 1, f[i][j]);

while (SPFA())
cost[++cnt] = augment();

rep (i, 1, cnt)     sum[i] = sum[i - 1] + cost[i];

int C, l, r;

while (Q--)
{
read(C);
l = 0, r = cnt;
while (l <= r)
{
int mid = (l + r) >> 1;
cost[mid] < C ? l = mid + 1 : r = mid - 1;
}
printf("%d\n", sum[r] + C * (n - r));
}

fclose(stdin);
fclose(stdout);

return 0;
}