pku 1275 Cashier Employment 差分约束

http://poj.org/problem?id=1275

题意：http://972169909-qq-com.iteye.com/blog/1185527 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;

}