# bzoj 2521 [Shoi2010]最小生成树

### Description

Secsa最近对最小生成树问题特别感兴趣。他已经知道如果要去求出一个n个点、m条边的无向图的最小生成树有一个Krustal算法和另一个Prim的算法。另外，他还知道，某一个图可能有多种不同的最小生成树。例如，下面图 3中所示的都是图 2中的无向图的最小生成树：

4 6 1

1 2 2

1 3 2

1 4 3

2 3 2

2 4 4

3 4 5

1

### HINT

1<=n<=500,1<=M<=800,1<=D< 10^6

### 题解

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define M 20100
#define N 10100
using namespace std;
const int inf = 0x3f3f3f3f;
struct flows {
struct node {
int to, next, flow;
} e[M];
int tot, st[N], dis[N], cur[N];
void init() {
tot = -1, memset(e, -1, sizeof e),
memset(st, -1, sizeof st);
}
void add(int x, int y, int z) {
e[++tot].to = y;
e[tot].flow = z;
e[tot].next = st[x];
st[x] = tot;
}
void add_edge(int x, int y, int z) {
//      printf("%d %d %d\n", x, y, z);
}
queue <int> que;
int bfs(int S, int T) {
memcpy(cur, st, sizeof st);
memset(dis, 0, sizeof dis);
while (!que.empty()) que.pop();
que.push(S);
dis[S] = 1;
while(!que.empty()) {
int now = que.front();
que.pop();
for (int i = st[now]; ~i; i = e[i].next)
if (e[i].flow && !dis[e[i].to]) {
dis[e[i].to] = dis[now] + 1;
if (e[i].to == T) return 1;
que.push(e[i].to);
}
}
return 0;
}
int finds(int now, int T, int flow) {
if (now == T)
return flow;
int f;
for (int i = cur[now]; ~i; i = e[i].next) {
cur[now] = i;
if (e[i].flow && dis[e[i].to] == dis[now] + 1 &&
(f = finds(e[i].to, T, min(flow, e[i].flow)))) {
e[i].flow -= f;
e[i^1].flow += f;
return f;
}
}
return 0;
}
int dinic(int S, int T) {
int flow = 0, x;
while(bfs(S, T)) {
while(x = finds(S, T, inf)) {
flow += x;
}
}
return flow;
}
}flow;
struct edge {
int to, next, val, from;
}e[M];
int st[N], tot;
void add(int x, int y, int z) {
e[++tot].to = y;
e[tot].next = st[x];
e[tot].from = x;
e[tot].val = z;
st[x] = tot;
}
bool comp(edge a, edge b) {
return a.val < b.val;
}
int n, m, t, fr, to, va, x, y, z;
main() {
scanf("%d%d%d", &n, &m, &t);
for (int i = 1; i <= m; ++i)
scanf("%d%d%d", &x, &y, &z),
flow.init();
for (int i = 1; i <= tot; ++i)
if (e[i].val <= e[t].val && i != t)
flow.add_edge(e[i].from, e[i].to, e[t].val - e[i].val + 1);
int S = e[t].from, T = e[t].to;
printf("%d\n", flow.dinic(S, T));
}