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

ACM模版

描述

题解

原题，直接拷贝代码，改都不用改，就能 AC。

求最小割边数。

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;}