zoj 1455 Schedule Problem 差分约束

设d[i]为 第i个工作开始的时间，根据题意有：

FAS a b: d[a] + time[a] >= d[b]

FAF a b: d[a] + time[a] >= d[b] + time[b]

SAF a b: d[a] >= d[b] + time[b]

SAS a b: d[a] >= d[b]

还有： d[i] >= 0 即 d[i] – d[0] >= 0

以0为源点开始搜索，0是超源点

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

const int MAX = 100000;
const int N = 10000;
const int INF = 1000000000;

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

Node node[MAX];
int cnt
;
int adj
;
bool in_q
;
int d
, times
;
int n;
int size;

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()
{
queue<int> Q;

memset(cnt, 0, sizeof(cnt));
memset(in_q, false, sizeof(in_q));
for (int i = 0; i <= n; i++)
d[i] = -INF;

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

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 (d[v] < d[u] + w)
{
d[v] = d[u] + w;
if (!in_q[v])
{
in_q[v] = true;
Q.push(v);

if (++cnt[v] > n) return false;
}
}
}
}

return true;
}

int main()
{
char c[20];
int a, b;
int cases = 1;

while (scanf("%d", &n) != EOF)
{
if (n == 0) break;

size = 0;
for (int i = 0; i <= n; i++)
adj[i] = -1;

for (int i = 1; i <= n; i++)
scanf("%d", ×[i]);

while (1)
{
getchar();
scanf("%s", c);
if (strcmp(c, "#") == 0)
break;
else if (strcmp(c, "FAS") == 0)
{
scanf("%d%d", &a, &b);
add_edge(b,a, -times[a]);
}
else if (strcmp(c, "FAF") == 0)
{
scanf("%d%d", &a, &b);
add_edge(b,a, times[b]-times[a]);
}
else if (strcmp(c, "SAF") == 0)
{
scanf("%d%d", &a, &b);
add_edge(b, a, times[b]);
}
else if (strcmp(c, "SAS") == 0)
{
scanf("%d%d", &a, &b);
add_edge(b, a, 0);
}
}

for (int i = 1; i <= n; i++)
add_edge(0, i, 0);
printf("Case %d:\n", cases++);

if (spfa())
{
for (int i = 1; i <= n; i++)
printf("%d %d\n", i, d[i]);
}
else
printf("impossible\n");

printf("\n");

}
return 0;
}