POJ 3189 枚举+最大流

注意到题中是所有牛中最高选择的等级和所有牛中最低等级之差最小

就可以想到枚举区间长度 然后 枚举起点 来进行最大流了

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 1111
#define MAXM 55555
#define INF 10000007
using namespace std;
struct node
{
    int v;    // vtex
    int c;    // cacity
    int f;   // current f in this arc
    int next, r;
}edge[MAXM];
int dist[MAXN], nm[MAXN], src, des, n;
int head[MAXN], e;
void add(int x, int y, int c)
{
    edge[e].v = y;
    edge[e].c = c;
    edge[e].f = 0;
    edge[e].r = e + 1;
    edge[e].next = head[x];
    head[x] = e++;
    edge[e].v = x;
    edge[e].c = 0;
    edge[e].f = 0;
    edge[e].r = e - 1;
    edge[e].next = head[y];
    head[y] = e++;
}
void rev_BFS()
{
    int Q[MAXN], h = 0, t = 0;
    for(int i = 1; i <= n; ++i)
    {
        dist[i] = MAXN;
        nm[i] = 0;
    }
    Q[t++] = des;
    dist[des] = 0;
    nm[0] = 1;
    while(h != t)
    {
        int v = Q[h++];
        for(int i = head[v]; i != -1; i = edge[i].next)
        {
            if(edge[edge[i].r].c == 0 || dist[edge[i].v] < MAXN)continue;
            dist[edge[i].v] = dist[v] + 1;
            ++nm[dist[edge[i].v]];
            Q[t++] = edge[i].v;
        }
    }
}
void init()
{
    e = 0;
    memset(head, -1, sizeof(head));
}
int maxflow()
{
    rev_BFS();
    int u;
    int total = 0;
    int cur[MAXN], rpath[MAXN];
    for(int i = 1; i <= n; ++i)cur[i] = head[i];
    u = src;
    while(dist[src] < n)
    {
        if(u == des)     // find an augmenting path
        {
            int tf = INF;
            for(int i = src; i != des; i = edge[cur[i]].v)
                tf = min(tf, edge[cur[i]].c);
            for(int i = src; i != des; i = edge[cur[i]].v)
            {
                edge[cur[i]].c -= tf;
                edge[edge[cur[i]].r].c += tf;
                edge[cur[i]].f += tf;
                edge[edge[cur[i]].r].f -= tf;
            }
            total += tf;
            u = src;
        }
        int i;
        for(i = cur[u]; i != -1; i = edge[i].next)
            if(edge[i].c > 0 && dist[u] == dist[edge[i].v] + 1)break;
        if(i != -1)     // find an admissible arc, then Advance
        {
            cur[u] = i;
            rpath[edge[i].v] = edge[i].r;
            u = edge[i].v;
        }
        else        // no admissible arc, then relabel this vtex
        {
            if(0 == (--nm[dist[u]]))break;    // GAP cut, Important!
            cur[u] = head[u];
            int mindist = n;
            for(int j = head[u]; j != -1; j = edge[j].next)
                if(edge[j].c > 0)mindist = min(mindist, dist[edge[j].v]);
            dist[u] = mindist + 1;
            ++nm[dist[u]];
            if(u != src)
                u = edge[rpath[u]].v;    // Backtrack
        }
    }
    return total;
}
int nt, b;
int val[MAXN][33];
int cc[33];
bool solve(int mid)
{
    src = nt + b + 1;
    des = nt + b + 2;
    n = des;
    for(int k = 1; k <= b - mid + 1; k++)
    {
        init();
        for(int i = 1; i <= nt; i++) add(src, i, 1);
        for(int i = 1; i <= b; i++) add(nt + i, des, cc[i]);
        for(int i = 1; i <= nt; i++)
            for(int j = k; j < k + mid; j++)
                add(i, val[i][j] + nt, 1);
        if(maxflow() == nt) return true;
    }
    return false;
}
int main()
{
    scanf("%d%d", &nt, &b);
    for(int i = 1; i <= nt; i++)
        for(int j = 1; j <= b; j++)
            scanf("%d", &val[i][j]);
    for(int i = 1; i <= b; i++) scanf("%d", &cc[i]);
    int low = 1, high = b;
    int ans = b;
    while(low <= high)
    {
        int mid = (low + high) >> 1;
        if(solve(mid))
        {
            ans = min(ans, mid);
            high = mid - 1;
        }
        else low = mid + 1;
    }
    printf("%d\n", ans);
    return 0;
}