P1343 地震逃生

题目描述
汶川地震发生时，四川**中学正在上课，一看地震发生，老师们立刻带领x名学生逃跑，整个学校可以抽象地看成一个有向图，图中有n个点，m条边。1号点为教室，n号点为安全地带，每条边都只能容纳一定量的学生，超过楼就要倒塌，由于人数太多，校长决定让同学们分成几批逃生，只有第一批学生全部逃生完毕后，第二批学生才能从1号点出发逃生，现在请你帮校长算算，每批最多能运出多少个学生，x名学生分几批才能运完。

输入格式
第一行3个整数n,m,x(x<2^31,n<=200,m<=2000)；以下m行，每行三个整数a,b,c（a1,a<>b,0描述一条边，分别代表从a点到b点有一条边，且可容纳c名学生。

输出格式两个整数，分别表示每批最多能运出多少个学生，x名学生分几批才能运完。如果无法到达目的地（n号点）则输出“Orz Ni Jinan
Saint Cow!”

输入输出样例

说明/提示【注释】

比如有图

1 2 100

2 3 1

100个学生先冲到2号点，然后1个1个慢慢沿2-3边走过去

18神牛规定这样是不可以的……

也就是说，每批学生必须同时从起点出发，并且同时到达终点

思路

Edmods_kraps / Dinic / ISAP

题解一

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

#define INF 0x3f3f3f3f
const int maxn = 205;
const int maxm = 2005;
struct Edge
{
    int v,w,next;
} edge[2*maxm];

int head[maxn];
int dis[maxn];      //每次bfs 都找最短路经 进行开刀
int use[maxn];
int pre[2*maxm];    //前驱数组，⚠️ pre[x] = i ,表示的意思是，到i这个节点的上一条边是 i
int min_flow[maxn];
int n, m, x;
int k = -1;

void Add(int u, int v, int w)
{
    edge[++ k] = (Edge){ v, w, head[u]}; head[u] = k;
    edge[++ k] = (Edge){ u, 0, head[v]}; head[v] = k;
}

//bfs用于分层
bool bfs(int s, int e)
{
    for(int i = 1; i <= n; i ++)
        dis[i] = INF, use[i] = 0, min_flow[i] = INF;
    dis[s] = 0;
    queue<int> q;
    q.push(s);
    int u,v;
    while(! q.empty())
    {
        u = q.front(); q.pop();
        use[u] = 0;

        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].v;
            if(dis[v] > dis[u] + 1 && edge[i].w)
            {
                dis[v] = dis[u] + 1;
                pre[v] = i;
                min_flow[v] = min(min_flow[u], edge[i].w);
                if(! use[v])
                {
                    q.push(v);
                    use[v] = 1;
                }
            }
        }
        if(min_flow[e] != INF)
            break;
    }
    if(min_flow[e] == INF)
        return false;
    //更新网络中的参与流量
    int now = e;
    while(now != s)
    {
        edge[pre[now]].w -= min_flow[e];
        edge[pre[now]^1].w += min_flow[e];
        now = edge[pre[now]^1].v;
    }
    return true;
}

int Edmonds_krap(int s, int e)
{
    int mx_flw = 0;
    while(bfs(s, e))
    {
        mx_flw += min_flow[e];
    }
    return mx_flw;
}

inline void init()
{
    for(int i = 0; i <= n; i ++)
        head[i] = -1;
    k = -1;

}

int main()
{
    ios::sync_with_stdio(false); cin.tie(0);
    //freopen("T.txt","r",stdin);
    cin >> n >> m >> x;
    init();
    int u, v, w;
    for(int i = 1; i <= m; i ++)
        cin >> u >> v >> w, Add(u, v, w);
    int res = Edmonds_krap(1, n);
    if(! res)
        cout << "Orz Ni Jinan Saint Cow!\n";
    else
    {
        cout << res << " ";
        int tem = x;
        tem /= res;
        if(tem * res == x)
            cout << tem << endl;
        else
            cout << tem + 1 << endl;
    }
    return 0;
}

题解一

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

#define INF 0x3f3f3f3f
const int maxn = 205;
const int maxm = 2005;
struct Edge
{
    int v,w,next;
} edge[2*maxm];

int head[maxn],cur[maxn];
int deep[maxn];
int n, m, x;
int k = -1;

void Add(int u, int v, int w)
{
    edge[++ k] = (Edge){ v, w, head[u]}; head[u] = k;
    edge[++ k] = (Edge){ u, 0, head[v]}; head[v] = k;
}

//bfs用于分层
bool bfs(int s, int e)
{
    for(int i = 0; i <= n; i ++)
        cur[i] = head[i], deep[i] = INF;
    deep[s] = 0;
    queue<int> q;
    q.push(s);
    int u,v;
    while(! q.empty())
    {
        u = q.front(); q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].v;
            if(edge[i].w && deep[v] == INF)
            {
                deep[v] = deep[u] + 1;
                q.push(v);
            }
        }
        if(deep[e] != INF)
            return true;
    }
    return false;
}

//dfs 用于增广网
int dfs(int now, int e, int limit)
{
    if(! limit || now == e) return limit;

    int flow = 0, f;
    for(int i = cur[now]; i != -1; i = edge[i].next)
    {
        if(deep[edge[i].v] == deep[now] + 1 && (f = dfs( edge[i].v, e, min( limit, edge[i].w))))
        {
            flow += f;          //答案把这一份合理的流量加上去
            limit -= f;         //限制条件也要改变，，这个limit 可能被一个点的多条实用。。
            edge[i].w -= f;     //正逆向边也要相应的操作
            edge[i^1].w += f;
            cur[now] = i;       //当前弧优化
            if(limit == 0)
                break;
        }
    }
    return flow;
}

int Dinic(int s, int e)
{
    int mx_flw = 0;
    while(bfs(s, e))
        mx_flw += dfs(s, e, INF);
    return mx_flw;
}

inline void init()
{
    for(int i = 0; i <= n; i ++)
        head[i] = -1;
    k = -1;

}

int main()
{
    ios::sync_with_stdio(false); cin.tie(0);
    //freopen("T.txt","r",stdin);
    cin >> n >> m >> x;
    init();
    int u, v, w;
    for(int i = 1; i <= m; i ++)
        cin >> u >> v >> w, Add(u, v, w);
    int res = Dinic(1, n);
    if(! res)
        cout << "Orz Ni Jinan Saint Cow!\n";
    else
    {
        cout << res << " ";
        int tem = x;
        tem /= res;
        if(tem * res == x)
            cout << tem << endl;
        else
            cout << tem + 1 << endl;
    }
    return 0;
}

题解一

#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

#define INF 1e9
const int maxn = 10005;
const int maxm = 200005;
int n, m, s, e;
int k;

struct Edge
{
    int v,w,next;
} edge[2 * maxm];

int head[maxn],cur[maxn],deep[maxn];
int last[maxm];
int num[maxm];      // num 桶，用来统计每个分层的节点的数量

void Add(int u, int v, int w)
{
    edge[++ k] = (Edge){ v, w, head[u]};
    head[u] = k;
    edge[++ k] = (Edge){ u, 0, head[v]};
    head[v] = k;
}

//bfs 用于更新deep层
void bfs(int e)
{
//    for(int i = 0; i <= n; i ++)
//        cur[i] = head[i], deep[i] = n;
    for(int i = 0; i <= m; i ++) cur[i] = head[i];
    for(int i = 1; i <= n; i ++) deep[i] = n;
    deep[e] = 0;
    queue<int> q;
    q.push(e);
    int u, v, w;
    while(! q.empty())
    {
        u = q.front(); q.pop();

        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].v;
            if(edge[i^1].w && deep[v] == n)      //正图 边存在 且 v这个节点没有被求过
            {
                deep[v] = deep[u] + 1;
                q.push(v);
            }
        }
    }
}

int Add_flow(int s, int e)
{
    int ans = INF;
    int now = e;
    while(now != s)
    {
        ans = min(ans, edge[last[now]].w);
        now = edge[last[now]^1].v;
    }
    now = e;
    while(now != s)
    {
        edge[last[now]].w -= ans;
        edge[last[now]^1].w += ans;
        now = edge[last[now]^1].v;
    }
    return ans;
}

int isap(int s, int e)
{
    int now = s;    //从起点开始进行操作
    bfs(e);         //先找出来一条边 被操作的增光路
    for(int i = 1; i <= n; i ++) num[deep[i]] ++;
    int mx_flw = 0;
    while(deep[s] < n)
    {
        if(now == e)        //如果到达汇点直接增广，重新回到源点进行下一轮增广
        {
            mx_flw += Add_flow(s, e);
            now = s;
        }
        bool has_find = 0;
        for(int i = cur[now]; i != -1; i = edge[i].next)
        {
            if(edge[i].w && deep[now] == deep[edge[i].v] + 1)
            {
                has_find = 1;   //做标记已经找到一种可行路径
                cur[now] = i;     //优化当前弧
                now = edge[i].v;
                last[edge[i].v] = i;
                break;
            }
        }

        if(! has_find)
        {
            int minn = n - 1;
            for(int i = head[now]; i != -1; i = edge[i].next)
                if(edge[i].w)
                    minn = min(minn, deep[edge[i].v]);
            if( (-- num[deep[now]]) == 0) break;   //gap 优化出现了断层
            num[deep[now] = minn + 1] ++;
            cur[now] = head[now];
            if(now != s)
                now = edge[last[now]^1].v;
        }
    }
    return mx_flw;
}


void init()
{
    k = -1;
    memset(head, -1, sizeof(head));
}

int main()
{
    ios::sync_with_stdio(false); cin.tie(0);
    //freopen("T.txt","r",stdin);
    int st;
    cin >> n >> m >> st;
    init();
    int u,v,w;
    for(int i = 1; i <= m; i ++)
        cin >> u >> v >> w, Add(u, v, w);

    int cnt = isap(1, n);
    if(! cnt)
        cout <<"Orz Ni Jinan Saint Cow!\n";
    else
    {
        cout << cnt << " ";
        int tem = st;
        tem /= cnt;
        if(tem * cnt == st)
            cout << tem << endl;
        else
            cout << tem + 1 << endl;
    }

    return 0;
}

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P1343 地震逃生（最大流板题）

P1343 地震逃生

思路

题解一

题解一

题解一

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