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pku 1275 Cashier Employment 差分约束

2012-06-24 16:36 183 查看
http://poj.org/problem?id=1275

题意:/article/4494905.html 这个解题报告描述的相当详细了。就不多说了;

差分约束关键是找出约束条件,然后建图。最后就是套spfa或者bellman_ford的模板就是了;

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define maxn 25
using namespace std;

struct node
{
int v,w;
int next;
}g[10*maxn];
int cnt,pre[maxn],ind[maxn],dis[maxn];
bool inq[maxn];
int R[maxn],num[26];
const int inf = 9999999;

void init()
{
cnt = 0;
memset(pre,-1,sizeof(pre));
memset(ind,0,sizeof(ind));
memset(inq,false,sizeof(inq));
}
void add(int u,int v,int w)
{
g[cnt].v = v;
g[cnt].w = w;
g[cnt].next = pre[u];
pre[u] = cnt++;
}
bool spfa(int s)
{
int i;
queue<int>q;
for (i = 0; i < maxn; ++i) dis[i] = -inf;
dis[s] = 0;
q.push(s); inq[s] = true;
while (!q.empty())
{
int u = q.front(); q.pop();
if (++ind[u] > 24) return false;
inq[u] = false;
for (i = pre[u]; i != -1; i = g[i].next)
{
int v = g[i].v, w = g[i].w;
if (dis[v] < dis[u] + w)
{
dis[v] = dis[u] + w;
if (!inq[v])
{
inq[v] = true;
q.push(v);
}
}
}
}
return true;
}
int main()
{
int i,t,pos,n;
scanf("%d",&t);
while (t--)
{
memset(num,0,sizeof(num));
for (i = 1; i <= 24; ++i) scanf("%d",&R[i]);
scanf("%d",&n);
for (i = 0; i < n; ++i)
{
scanf("%d",&pos);
num[pos + 1]++;
}
int l = 0, r = n;
bool flag = false;
while (l < r)
{
init();
int mid = (l + r)>>1;
for (i = 1; i <= 24; ++i)
{
add(i - 1,i,0);
add(i,i - 1,-num[i]);
}
for (i = 8; i <= 24; ++i)
add(i - 8,i,R[i]);
for (i = 1; i <= 7; ++i)
add(i + 16,i,R[i] - mid);
add(0,24,mid);
if (spfa(0))
{
r = mid;
flag = true;
}
else
{
l = mid + 1;
}
}
if (flag) printf("%d\n",r);
else printf("No Solution\n");
}
return 0;
}


  
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