任重而道远
给定一个边带正权的连通无向图G=(V,E),其中N=|V|,M=|E|,N个点从1到N依次编号,给定三个正整数u,v,和L (u≠v),假设现在加入一条边权为L的边(u,v),那么需要删掉最少多少条边,才能够使得这条边既可能出现在最小生成树上,也可能出现在最大生成树上?
Input
第一行包含用空格隔开的两个整数,分别为N和M;
接下来M行,每行包含三个正整数u,v和w表示图G存在一条边权为w的边(u,v)。
最后一行包含用空格隔开的三个整数,分别为u,v,和 L;
数据保证图中没有自环。
Output
输出一行一个整数表示最少需要删掉的边的数量。
Sample Input
3 2 3 2 1 1 2 3 1 2 2
Sample Output
1
Hint
对于20%的数据满足N ≤ 10,M ≤ 20,L ≤ 20;
对于50%的数据满足N ≤ 300,M ≤ 3000,L ≤ 200;
对于100%的数据满足N ≤ 20000,M ≤ 200000,L ≤ 20000。
AC代码:
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int oo = 1e9 + 7;
const int N = 2e4 + 4;
const int M = 2e5 + 5;
struct Node {
int u, v, val;
bool operator < (const Node &a) const {
return val < a.val;
}
}g[M];
struct Edge {
int tov, nxt, val;
}e[M << 2];
int head[N], dis[N], vis[N];
int num = 1, n, m, src, sink, L;
long long sww;
void add_edge (int u, int v, int val) {
e[++num] = (Edge) {v, head[u], val}, head[u] = num;
}
bool bfs () {
memset (dis, 127 / 2, sizeof (dis));
memset (vis, 0, sizeof (vis));
queue <int> q;
q.push (src), dis[src] = 0, vis[src] = 1;
while (!q.empty ()) {
int u = q.front ();
q.pop ();
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].tov;
if (e[i].val && !vis[v]) {
dis[v] = dis[u] + 1;
vis[v] = 1;
q.push (v);
}
}
}
return dis[sink] < dis[0];
}
int dfs (int u, int delta) {
if (u == sink) return delta;
int res = 0;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].tov;
if (e[i].val && dis[v] == dis[u] + 1) {
int d = dfs (v, min (delta, e[i].val));
e[i].val -= d;
e[i ^ 1].val += d;
delta -= d;
res += d;
}
}
return res;
}
int main () {
scanf ("%d%d", &n, &m);
for (int i = 1; i <= m; i++) {
int u, v, val;
scanf ("%d%d%d", &u, &v, &val);
g[i] = (Node) {u, v, val};
}
sort (g + 1, g + m + 1);
scanf ("%d%d%d", &src, &sink, &L);
for (int i = 1; i <= m; i++) {
if (g[i].val >= L) break;
int u = g[i].u, v = g[i].v;
add_edge (u, v, 1), add_edge (v, u, 1);
}
while (bfs ()) sww += dfs (src, oo);
num = 1, memset (head, 0, sizeof (head));
for (int i = m; i >= 1; i--) {
if (g[i].val <= L) break;
int u = g[i].u, v = g[i].v;
add_edge (u, v, 1), add_edge (v, u, 1);
}
while (bfs ()) sww += dfs (src, oo);
printf ("%lld\n", sww);
return 0;
}