HDU-2017 ACM/ICPC Asia Regional Qingdao Online-1009-Smallest Minimum Cut

描述

题解

代码

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

typedef long long ll;

const int MAXN = 10000;
const ll INF = 1ll << 60;
const int MOD = 100001;
const int MAXM = 500000;

int n, m;
int level[MAXN], que[MAXN];
int head[MAXN], lon;

ll min(ll a, ll b)
{
if (a < b)
{
return a;
}
else
{
return b;
}
}

struct EDGE
{
int next, to;
ll c;
} e[MAXM];

void edgeini()
{
memset(head, -1, sizeof(head));
lon = -1;
}

void edgemake(int from, int to, ll c)
{
e[++lon].c = c;
e[lon].to = to;
e[lon].next = head[from];
head[from] = lon;
}

void make(int from, int to, ll c)
{
edgemake(from, to, c);
edgemake(to, from, 0);
}

bool makelevel(int s, int t)
{
memset(level, 0, sizeof(level));
int front = 1, end = 0;
que[++end] = s;
level[s] = 1;
while (front <= end)
{
int u = que[front++];
if (u == t)
{
return true;
}
for (int k = head[u]; k != -1; k = e[k].next)
{
int v = e[k].to;
if (!level[v] && e[k].c)
{
que[++end] = v;
level[v] = level[u] + 1;
}
}
}

return false;
}

ll dfs(int now, int t, ll maxf)
{
if (now == t || maxf == 0)
{
return maxf;
}
ll ret = 0;
for (int k = head[now]; k != -1; k = e[k].next)
{
int u = e[k].to;
if (level[u] == level[now] + 1 && e[k].c)
{
ll f = dfs(u, t, min(e[k].c, maxf - ret));
e[k].c -= f;
e[k^1].c += f;
ret += f;
if (ret == maxf)
{
return ret;
}
}
}
if (ret == 0)
{
level[now] = 0;
}
return ret;
}

ll maxflow(int s, int t)
{
ll ret = 0;
while (makelevel(s, t))
{
ret += dfs(s, t, INF);
}
return ret;
}

int main()
{
int cas;
scanf("%d", &cas);

int sum = 0;
int s, t;
int u, v;
ll w;

while (cas--)
{
sum++;

scanf("%d%d", &n, &m);
scanf("%d%d", &s, &t);

edgeini();

for (int i = 1; i <= m; i++)
{
scanf("%d%d%lld", &u, &v, &w);
make(u, v, w * MOD + 1);
}
printf("%lld\n", maxflow(s, t) % MOD);
}

return 0;
}