POJ-1275/HDU-1529 Cashier Employment

来源：互联网发布：spss 贝叶斯网络示例编辑：程序博客网时间：2024/05/04 07:53

题目大意：

从0点到23点，给出每个时刻需要的售货员个数，再给出每个时刻应征的售货员个数，然后让你求出满足需求的最小售货员个数

解题思路：

差分约束

代码：

#include <queue>#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;const int maxn = 1000;typedef struct node{    int to, w;    int next;}Edge;int tot;Edge edge[maxn];int r[25], t[25];int head[maxn], dis[30], vis[30], cnt[30];void add(int u, int v, int w){    edge[tot].w = w;    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}int spfa(int ans){    int p, v;    queue<int> q;    while(!q.empty()) q.pop();    for(int i = 0; i <= 24; ++i){        dis[i] = 0;        vis[i] = 1;        cnt[i] = 0;        q.push(i);    }    while(!q.empty()){        p = q.front(); q.pop(); vis[p] = 0;        if(++cnt[p] > 25) return 0;        for(int i = head[p]; ~i; i = edge[i].next){            v = edge[i].to;            if(dis[v] > dis[p] + edge[i].w){                dis[v] = dis[p] + edge[i].w;                if(!vis[v]){                    q.push(v);                    vis[v] = 1;                }            }        }    }    if(-dis[24] == ans) return 1;    else return 0;}int solve(int ans){    tot = 0;    memset(head, -1, sizeof(head));    for(int i = 0; i < 24; ++i){        add(i, i+1, 0);        add(i+1, i, t[i+1]);    }    for(int j = 1; j <= 24; ++j){        if(j + 8 <= 24) add(j, j + 8, -r[j+8]);        else add(j, j-16, ans - r[j-16]);    }    add(0, 24, -ans);    if(spfa(ans)) return 1;    else return 0;}int main(){    int tt, n, a;    scanf("%d", &tt);    while(tt--){        for(int i = 1; i <= 24; ++i){            scanf("%d", &r[i]); t[i] = 0;        }        scanf("%d", &n);        for(int i = 0; i < n; ++i){            scanf("%d", &a); ++t[a+1];        }        int flag = 0;        for(int i = 0; !flag && i <= n; ++i){            flag = solve(i);            if(flag) {printf("%d\n", i); break;}        }        if(!flag) puts("No Solution");    }    return 0;}

0 0