SPOJ 287 Smart Network Administrator

SPOJ_287

    YY了一下颜色的数量取决于网络流中各条边中最大的流量，于是二分边的容量并做网络流就可以了。

    但至于为什么颜色的数量取决于网络流中各条边中最大的流量，暂时没有细加证明……

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXD 510
#define MAXM 501010
#define INF 0x3f3f3f3f
int N, M, K, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM], list[MAXD];
int S, T, d[MAXD], q[MAXD], work[MAXD];
struct Edge
{
    int x, y;
}edge[MAXM];
void init()
{
    int i;
    scanf("%d%d%d", &N, &M, &K);
    for(i = 0; i < K; i ++)
        scanf("%d", &list[i]);
    for(i = 0; i < M; i ++)
        scanf("%d%d", &edge[i].x, &edge[i].y);
}
void add(int x, int y, int z)
{
    v[e] = y, flow[e] = z;
    next[e] = first[x], first[x] = e ++;    
}
void build(int c)
{
    int i;
    memset(first, -1, sizeof(first[0]) * (T + 1));
    e = 0;
    add(S, 1, K), add(1, S, 0);
    for(i = 0; i < K; i ++)
        add(list[i], T, 1), add(T, list[i], 0);
    for(i = 0; i < M; i ++)
        add(edge[i].x, edge[i].y, c), add(edge[i].y, edge[i].x, c);
}
int bfs()
{
    int i, j, rear = 0;
    memset(d, -1, sizeof(d[0]) * (T + 1));
    d[S] = 0, q[rear ++] = S;
    for(i = 0; i < rear; i ++)
        for(j = first[q[i]]; j != -1; j = next[j])
            if(flow[j] && d[v[j]] == -1)
            {
                d[v[j]] = d[q[i]] + 1, q[rear ++] = v[j];
                if(v[j] == T)
                    return 1;    
            }    
    return 0;
}
int dfs(int cur, int a)
{
    if(cur == T)
        return a;
    int t;
    for(int &i = work[cur]; i != -1; i = next[i])
        if(flow[i] && d[v[i]] == d[cur] + 1)
            if(t = dfs(v[i], std::min(a, flow[i])))
            {
                flow[i] -= t, flow[i ^ 1] += t;
                return t;    
            }
    return 0;
}
int dinic()
{
    int ans = 0, t;
    while(bfs())
    {
        memcpy(work, first, sizeof(first[0]) * (T + 1));
        while(t = dfs(S, INF))
            ans += t;
    }    
    return ans;
}
void solve()
{
    int min, max, mid;
    S = 0, T = N + 1;
    min = -1, max = K;
    for(;;)
    {
        mid = min + max + 1 >> 1;
        if(max == mid) break;
        build(mid);
        if(dinic() == K)
            max = mid;
        else
            min = mid;
    }
    printf("%d\n", mid);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        solve();    
    }
    return 0;
}