hdu 1529 差分约束(Bellman_Ford)

Problem : 1529 ( Cashier Employment )     Judge Status : Accepted
RunId : 1599582    Language : C++    Author : va
Code Render Status : Rendered By HDOJ C++ Code Rander Version 0.01 Beta

#include <cstdio>
#include <cstring>

const int MAXN = 1010;
typedef struct
{
    int from, to, cost;
}Edge;

int min_need[25]; //各时刻最少需要工作人数
int now_can[25];  //各时刻能*开始*工作人数
int dis[25];      //各时刻实际需要人数
Edge e[MAXN];
int n, t, oldt;

//初始化
void init()
{
    int i; 
    memset(now_can, 0, sizeof(now_can));
    for (i = 0; i < 24; ++i)
    {
        scanf("%d", &min_need[i]);
    }
    scanf("%d", &n);
    for (i = 0; i < n; ++i)
    {
        scanf("%d", &t);
        now_can[t]++;
    }
}

//建图
void construct_graphic()
{
    int i;
    t = 0;
    for (i = 1; i < 24; ++i)
    {
        e[t].from = i;
        e[t].to = i-1;
        e[t++].cost = 0;
    }
    for (i = 1; i < 24; ++i)
    {
        e[t].from = i-1;
        e[t].to = i;
        e[t++].cost = now_can[i];
    }
    for (i = 8; i < 24; ++i)
    {
        e[t].from = i;
        e[t].to = i-8;
        e[t++].cost = -min_need[i];
    }
    oldt = t;
}

bool bellman_ford(int ans)
{
    int i, j;
    t = oldt;
    for (i = 0; i <= 7; ++i)
    {
        e[t].from = i;
        e[t].to = i+16;
        e[t++].cost = ans - min_need[i];
    }
    memset(dis, 0, sizeof(dis));//赋0 即构造新的不等式组xi-x0<=0 x0为起点
    bool flag = true;
    for (i = 0; i <= 25 && flag; ++i) //多了1次循环 是为了判负圈的情况
    {
        flag = false;
        for (j = 0; j < t; ++j)
        {
            if (dis[e[j].from] + e[j].cost < dis[e[j].to])
            {
                dis[e[j].to] = dis[e[j].from] + e[j].cost;
                flag = true;
            }
        }
    }
    if (flag)
    {
        return false;
    }
    else
    {
        return true;
    }
}

//对s[23]二分
void binary_search()
{
    int high, low, mid, ans = -1;
    high = n, low = 0;
    while (low < high)
    {
        mid = (low + high)/2;
        if (bellman_ford(mid))
        {
            ans = mid;
            high = mid;
        }
        else
        {
            low = mid+1;
        }
    }
    if (ans == -1)
    {
        puts("No Solution");
    }
    else
    {
        printf("%d\n", ans);
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        init();
        construct_graphic();
        binary_search();
    }
    return 0;
}