poj 1273 Drainage Ditches 最大流

解题思路: 源点为1,汇点为N, 直接求最大流就好

View Code

#include<stdio.h>

#include<string.h>

#include<string.h>

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

const int inf = ~0u>>1;

const int MAXN = 210;

int n, m, S, T, N;

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

struct Edge{

    int v, f, nxt;    

}edge[MAXN<<4];



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 build()

{

    int u, v, f;

    S = 1; T = n; N = n;    

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

    idx = 0;

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

    {

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

        AddEdge( u, v, f );

    }

}

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;

}

void sap()

{

    int maxflow = 0;

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

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

    vh[0] = N;

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

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

}

int main()

{

    while( scanf("%d%d", &m,&n) != EOF)

    {

        build();

        sap();

    }

    return 0; 

}