牛棚安排
gmoj 1259
題目大意:
有頭牛和個牛棚,每頭牛有自己第1喜歡,第2喜歡……第喜歡的牛棚(開心度分別爲),然後讀入個數表示第個牛棚最多可以進多少頭牛,現在讓所有牛都進入到牛棚內,問最開心的牛和最傷心的牛開心度之差最小是多少
輸入樣例
6 4
1 2 3 4
2 3 1 4
4 2 3 1
3 1 2 4
1 3 4 2
1 4 2 3
2 1 3 2
輸出樣例
2
樣例解釋
每頭奶牛都能被安排進她的第一或第二喜歡的牛棚。下面給出一種合理的分配方案:奶牛1和奶牛5住入牛棚1,牛棚2由奶牛2獨佔,奶牛4住進牛棚3,剩下的奶牛3和奶牛6安排到牛棚4。
解題思路
先二分枚舉答案,然後枚舉每一個區間,接下來用最大流求二分圖的匹配即可
代碼
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
int n, m, l, r, mid, sum, s, t, tot, b[50], head[2000], dep[2000], v[2100][50];
const int inf = 1 << 29;
struct rec
{
int to, next, edge;
}a[40500];
queue<int> d;
void add(int x, int y, int z)
{
a[++tot].to = y;
a[tot].edge = z;
a[tot].next = head[x];
head[x] = tot;
a[++tot].to = x;
a[tot].edge = 0;
a[tot].next = head[y];
head[y] = tot;
}
bool bfs()
{
memset(dep, 0, sizeof(dep));
while(!d.empty()) d.pop();
d.push(s);
dep[s] = 1;
while (!d.empty())
{
int h = d.front();
d.pop();
for (int i = head[h]; i; i = a[i].next)
if (!dep[a[i].to] && a[i].edge)
{
dep[a[i].to] = dep[h] + 1;
if (a[i].to == t) return true;
d.push(a[i].to);
}
}
return false;
}
int dinic(int x, int flow)
{
if (x == t) return flow;
int rest = 0, k;
for (int i = head[x]; i; i = a[i].next)
if (dep[x] + 1 == dep[a[i].to] && a[i].edge)
{
k = dinic(a[i].to, min(a[i].edge, flow - rest));
if (!k) dep[a[i].to] = 0;
rest += k;
a[i].edge -= k;
a[i^1].edge += k;
if (rest == flow) return rest;
}
return rest;
}
bool js(int x)
{
s = 0;
t = n + m + 1;
for (int i = 1; i <= m - x + 1; ++i)//開頭位置
{
memset(head, 0, sizeof(head));
tot = 0;
for (int j = 1; j <= m; ++j)
add(n + j, t, b[j]);//連到T
for (int j = 1; j <= n; ++j)
{
add(s, j, 1);//連到S
for (int k = i; k - i + 1 <= x; ++k)
add(j, n + v[j][k], 1);//連邊
}
sum = 0;
while(bfs())//找增廣路
sum += dinic(s, inf);//流
if (sum == n) return true;//可以滿足
}
return false;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j)
scanf("%d", &v[i][j]);//先存下來,到時再連邊
for (int i = 1; i <= m; ++i)
scanf("%d", &b[i]);
l = 1;
r = m;
while(l < r)//二分枚舉
{
mid = (l + r) >> 1;
if (js(mid)) r = mid;
else l = mid + 1;
}
printf("%d", l);
return 0;
}