[BZOJ2654][tree][二分+Kruskal]

[BZOJ2654][tree][二分+Kruskal]

题目大意：

给你一个无向带权连通图，每条边是黑色或白色。让你求一棵最小权的恰好有need条白色边的生成树。

思路：

可以给每条白色边都加一个权值，这个权值越大，选的白边就越少，反之就越多。对于这个权值二分就好了，然后再做一遍最小生成树。

代码：

#include <bits/stdc++.h>
using namespace std;
const int Maxn = 100010, INF = 200;
struct Edge {
int u, v, x, c;
} edge[Maxn];
int fa[Maxn], uu[Maxn], vv[Maxn], c[Maxn], xx[Maxn];
int n, m, ned;
inline bool cmp(Edge a, Edge b) {
return a.x == b.x ? a.c < b.c : a.x < b.x;
}
inline int getf(int x) {
return fa[x] = (x == fa[x] ? x : getf(fa[x]));
}
inline void read(int &x) {
x = 0; static char c;
for (; !(c >= '0' && c <= '9'); c = getchar());
for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = getchar());
}
int val, chosen;
inline bool check(int x) {
for (int i = 0; i < m; i++) {
edge[i].u = uu[i], edge[i].v = vv[i], edge[i].c = c[i], edge[i].x = xx[i];
if (!c[i]) edge[i].x += x;
}
sort(edge, edge + m, cmp);
for (int i = 1; i <= n; i++) fa[i] = i;
val = chosen = 0;
int i = 0, fu, fv;
while (i < m) {
fu = getf(edge[i].u), fv = getf(edge[i].v);
if (fu != fv) {
val += edge[i].x;
if (!edge[i].c) ++chosen;
fa[fu] = fv;
}
i++;
}
return chosen >= ned;
}
int main(void) {
read(n), read(m), read(ned);
for (int i = 0; i < m; i++) {
read(uu[i]), read(vv[i]), read(xx[i]), read(c[i]);
uu[i]++, vv[i]++;
}
int L = -INF, R = INF, mid, total;
while (L <= R) {
mid = (L + R) >> 1;
if (check(mid)) L = mid + 1, total = val - mid * ned;
else R = mid - 1;
}
cout << total << endl;
return 0;
}

完。

By g1n0st