poj 1275 Cashier Employment 差分约束

这题的不等式比较难列，列了很久，也列不全，对那个8小时理解的不透。

设num[i] 为来应聘的在第i个小时开始工作的人数

r[i] 为第i个小时至少需要的人数

x[i] 为招到的在第i个小时开始工作的人数

根据题意有：

0 <= x[i] <= num[i]

x[i] + x[i-1] + …+ x[i-7] >= r[i] (题目中的连续工作8小时)

再设 s[i] = x[1] + … + x[i]

则有： s[i] – s[i-1] >= 0

s[i-1] – s[i] >= –num[i]

s[i] – s[i-8] >= r[i], 8 <= i <= 24

s[i] – s[i+16] >= r[i] – s[24], 1<= i <= 7

还需要添加一个隐藏不等式： s[24] – s[0] >= ans（枚举的答案）

通过枚举s[24],来检测是否满足条件，题目是求最小值，即求最长路，以0为源点。

#include <iostream>
#include <queue>
using namespace std;

const int N = 30;
const int MAX = 10000;
const int INF = 1000000000;

struct Node
{
int v;
int cost;
int next;
};

Node node[MAX];
queue<int> Q;
int adj
;
int num
;
int size;
int r
;
int s
;
int cnt
;
bool in_q
;

void add_edge(int u, int v, int cost)
{
node[size].v = v;
node[size].cost = cost;
node[size].next = adj[u];
adj[u] = size++;
}

bool spfa(int ans)
{
for (int i = 0; i <= 24; i++)
{
s[i] = -INF;
cnt[i] = 0;
in_q[i] = false;
}

while (!Q.empty()) Q.pop();

int u, v, w;
s[0] = 0;
Q.push(0);
in_q[0] = true;

while (!Q.empty())
{
u = Q.front();
Q.pop();
in_q[u] = false;

for (int i = adj[u]; i != -1; i = node[i].next)
{
v = node[i].v;
w = node[i].cost;

if (s[v] < s[u] + w)
{
s[v] = s[u] + w;

if (!in_q[v])
{
in_q[v] = true;
Q.push(v);
if (++cnt[v] > 24)
return false;
}
}
}
}
if (s[24] == ans)
return true;
else
return false;
}
int main()
{
int cases;
int n, c;
bool has_solution;

cin >> cases;

while (cases--)
{
for (int i = 1; i <= 24; i++)
{
scanf("%d", &r[i]);
num[i] = 0;
}

cin >> n;
for (int i = 0; i < n; i++)
{
scanf("%d", &c);
num[c+1]++;
}
int ans;

for ( ans = 0; ans <= n; ans++)
{
size = 0;
has_solution = false;
for (int i = 0; i <= 24; i++)
adj[i] = -1;

for (int i = 1; i <= 24; i++)
{
add_edge(i-1, i, 0);
add_edge(i, i-1, -num[i]);

if (i >= 8)
add_edge(i-8, i, r[i]);
}

for (int i = 1; i <= 7; i++)
add_edge(i+16, i, r[i]- ans);

add_edge(0, 24, ans);

if (spfa(ans))
{
has_solution = true;
break;
}

}

if (has_solution)
printf("%d\n", ans);
else
printf("No Solution\n");
}
return 0;
}