2011清华集训.BZOJ2561 && THU A1277.最小生成树(最小割)

有一个 n 个点， m 条边的图，现在多加入一条边 <u, v>，求最少要删去多少条边，才能使边
<u, v> 既可能出现在最小生成树上，也可能出现在最大生成树上。

设后加入的边的长度为 L，那么只有原图中小于 L
的边不能再用的时候，<u, v> 才可能出现在最小生成树上，用最小割可求。最大生成树时同理。两次答案相加。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX_N = 20005, MAX_M = 200005;
const int inf = 1000000007;

struct node {
int v, w, next;
}E[MAX_M << 1];
struct arr {
int x, y, k;
}R[MAX_M];
int head[MAX_N], cur[MAX_N], top = 0;
int n, m, s, t, l, ans = 0;
int dep[MAX_N], q[MAX_M];

inline void add(int u, int v, int x) {
int t1 = x << 1, t2 = t1 | 1;
E[t1].v = v; E[t1].w = 1; E[t1].next = head[u]; head[u] = t1;
E[t2].v = u; E[t2].w = 1; E[t2].next = head[v]; head[v] = t2;
}
inline bool cmp(arr a, arr b) {
return a.k < b.k;
}
inline bool bfs() {
memset(dep, -1, sizeof(dep));
int bg = 0, ed = 1;
q[ed] = s; dep[s] = 0;
while (bg < ed) {
int x = q[++ bg];
for (int i = head[x]; i != -1; i = E[i].next)
if (dep[E[i].v] == -1 && E[i].w) {
dep[E[i].v] = dep[x] + 1; q[++ ed] = E[i].v;
if (E[i].v == t) return 1;
}
}
return 0;
}
int dfs(int x, int mx) {
if (x == t) return mx;
int ret;
for (int &i = cur[x]; i != -1; i = E[i].next)
if (E[i].w && dep[E[i].v] == dep[x] + 1 && (ret = dfs(E[i].v, min(mx, E[i].w)))) {
E[i].w -= ret; E[i ^ 1].w += ret;
return ret;
}
return 0;
}
void dinic() {
int ret;
while (bfs()) {
for (int i = 0; i <= n; i ++) cur[i] = head[i];
while (ret = dfs(s, inf)) ans += ret;
}
}
void init() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= m; i ++)
scanf("%d%d%d", &R[i].x, &R[i].y, &R[i].k);
scanf("%d%d%d", &s, &t, &l);
}
void doit() {
sort(R + 1, R + 1 + m, cmp);
memset(head, -1, sizeof(head)); top = 0;
for (int i = 1; i <= m; i ++) {
if (R[i].k < l) add(R[i].x, R[i].y, top ++);
else break;
}
dinic();
memset(head, -1, sizeof(head)); top = 0;
for (int i = m; i >= 1; i --) {
if (R[i].k > l) add(R[i].x, R[i].y, top ++);
else break;
}
dinic();
printf("%d\n", ans);
}
int main() {
init();
doit();
return 0;
}