WZRYWZWY @ 2024-02-23 08:11:39
rt,按着《算法竞赛进阶指南》打的,感觉一模一样QAQ
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int inf = 1 << 31, N = 205, M = 10005;
int head[N], ver[M], edge[M], Next[M], d[N], /*当前弧优化*/now[M];
int n, m, s, t, tot, maxflow;
queue <int> q;
void add(int x, int y, int z) {
ver[++tot] = y, edge[tot] = z, Next[tot] = head[x], head[x] = tot;
ver[++tot] = x, edge[tot] = 0, Next[tot] = head[y], head[y] = tot;
}
bool bfs() { // 在残量网络上构造分层图
memset(d, 0, sizeof(d));
while (q.size()) q.pop();
q.push(s); d[s] = 1; now[s] = head[s];
while (q.size()) {
int x = q.front(); q.pop();
for (int i = head[x]; i; i = Next[i])
if (edge[i] && !d[ver[i]]) {
q.push(ver[i]);
now[ver[i]] = head[ver[i]];
d[ver[i]] = d[x] + 1;
if (ver[i] == t) return 1;
}
}
return 0;
}
int dinic(int x, int flow) { // 在当前分层图上增广
if (x == t) return flow;
int rest = flow, k, i;
for (i = now[x]; i && rest; i = Next[i]) {
now[x] = i; // 当前弧优化(避免重复遍历从 x 出发不可扩展的边)
if (edge[i] && d[ver[i]] == d[x] + 1) {
k = dinic(ver[i], min(rest, edge[i]));
if (!k) d[ver[i]] = 0; // 剪枝,去掉增广完毕的点
edge[i] -= k;
edge[i ^ 1] += k;
rest -= k;
}
}
return flow - rest;
}
signed main() {
cin >> n >> m >> s >> t;
tot = 1;
for (int i = 1; i <= m; i++) {
int x, y, c; cin >> x >> y >> c;
add(x, y, c);
}
int flow = 0;
while (bfs())
while (flow = dinic(s, inf)) maxflow += flow;
cout << maxflow;
}
by ___Furina___ @ 2024-02-23 08:20:47
@WZRYWZWY 这一句话错了(笑):
const int inf = 1 << 31, N = 205, M = 10005;
应该改成:
const int inf = 2e9, N = 205, M = 10005;
by QWQ_123 @ 2024-02-23 08:26:01
@WZRYWZWY 1 << 31
太大了,改成 1 << 30
即可。
by WZRYWZWY @ 2024-02-23 08:31:19
@OIer_qyzy @QWQ_123 thanks.
by WZRYWZWY @ 2024-02-23 08:35:44
@OIer_qyzy @QWQ_123 嗯,但是不是
by ___Furina___ @ 2024-02-23 08:38:26
@WZRYWZWY
by ___Furina___ @ 2024-02-23 08:40:35
@WZRYWZWY 德莉傻病毒已经入侵 OI 了吗?qnq
by WZRYWZWY @ 2024-02-23 08:45:17
@OIer_qyzy 不好意思打错数据范围了,我主要是想表达31次方((
话说为什么还是int的大小
by WZRYWZWY @ 2024-02-23 08:56:39
@QWQ_123 @OIer_qyzy
找到问题了,应该是
此贴结.