2013多校第四场 G题 ZZ的搬砖难题

虽然题目只给了起点st,和终点ed,  st,ed <= 10000.

但是只有200条边,极端情况也才200条边对应的400个顶点都不一样.

所以我们可以离散化顶点到[1,400]之间.然后跑个最大流即可.

注意本题边是单向的.

View Code

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<map>

using namespace std;

const int inf = 0x3f3f3f3f;

const int MAXN = 1e5+10;

#define MIN(a,b) (a)<(b)?(a):(b)

#define MAX(a,b) (a)>(b)?(a):(b)

int k, c, m, s, t, n, N;

int head[MAXN], idx, vh[MAXN], h[MAXN];

struct node{

    int v, f, nxt;

}edge[MAXN];



struct Tmp{

    int u, v, c;

    void input(){

        scanf("%d%d%d", &u,&v,&c);    

    }    

}test[210];

map<int,int> mp;

int st, ed;



void AddEdge( int u, int v, int f )

{

    edge[idx].v = v; edge[idx].f = f;

    edge[idx].nxt = head[u]; head[u] = idx++;

    edge[idx].v = u; edge[idx].f = 0;

    edge[idx].nxt = head[v]; head[v] = idx++;

}



void input() 

{

    scanf("%d", &m);

    int cnt = 0;

    mp.clear();    

    if( mp.count(st) == 0 ) mp[st] = ++cnt;

    if( mp.count(ed) == 0 ) mp[ed] = ++cnt;

    for(int i = 0; i < m; i++){

        test[i].input();

        if( mp.count( test[i].u ) == 0 ) mp[ test[i].u ] = ++cnt;

        if( mp.count( test[i].v ) == 0 ) mp[ test[i].v ] = ++cnt;

    }

    N = cnt;

    memset( head, 0xff, sizeof(head) ); idx = 0;

    s = mp[st]; t = mp[ed];

    for(int i = 0; i < m; i++){

        AddEdge( mp[test[i].u], mp[test[i].v], test[i].c );

        AddEdge( mp[test[i].v], mp[test[i].u], test[i].c );

    }

//    AddEdge( s, mp[st], inf );

//    AddEdge( mp[ed], t, inf );

}

int DFS(int u,int flow )

{

    if( u == t ) return flow;

    int tmp = h[u]+1, remain = flow;

    for(int i = head[u]; ~i; i = edge[i].nxt )

    {

        int v = edge[i].v;

        if( edge[i].f && h[u] == h[v]+1 )

        {

            int p = DFS( v, MIN( remain, edge[i].f ));

            edge[i].f -= p; edge[i^1].f += p; remain -= p;

            if( remain == 0 || h[s] == N ) return flow-remain;

        }

    }

    for(int i = head[u]; ~i; i = edge[i].nxt )

        if( edge[i].f ) tmp = MIN( tmp, h[ edge[i].v ] );

    if( !( --vh[ h[u] ] ) ) h[s] = N;

    else    ++vh[ h[u] = tmp+1 ];

    return flow-remain;

}

int sap()

{

    int maxflow = 0;

    memset( h, 0, sizeof(h));

    memset( vh, 0, sizeof(vh));

    vh[0] = N;

    while( h[s] < N ) maxflow += DFS( s, inf );

    return maxflow;

}

int main()

{

    while( scanf("%d%d", &st, &ed ) != EOF)

    {

        input();

        int res = sap();    

        printf("%d\n", res ); 

    }

    return 0;

}