|hdu 3440|差分约束|House Man

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hdu 3440

差分约束，由于求最大差，故建立a−b<=c的不等式，跑最短路
h数组排序后，依次连接两个相邻的数，注意绝对值的问题

ps: INF要开0x7fffffff

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<queue>#define ms(i, j) memset(i, j, sizeof i)#define FN2 "hdu3440" using namespace std;const int MAXN = 1000 + 5, ZINF2 = 0x7fffffff;struct edge{    int u, v, c;}e[MAXN*MAXN];struct house{    int no, hi;    bool operator < (const house b) const     {        return hi < b.hi;    }}h[MAXN];struct sdc{    vector<int> G[MAXN];    int dis[MAXN], cir[MAXN], vi[MAXN], n, en;    void init(int n)    {        en = 0;        this->n = n;        for (int i=0;i<=n;i++) G[i].clear(), dis[i] = ZINF2, cir[i] = vi[i] = 0;    }    void addedge(int a, int b, int c)//a-b<=c    {        en++; e[en].u = b, e[en].v = a, e[en].c = c;        G[b].push_back(en);    }    int solve(int from, int to)    {        queue<int> q;        cir[from] = 1, vi[from] = true, dis[from] = 0, q.push(from);        while(!q.empty())        {            int p = q.front(); q.pop(); vi[p] = false;            for (int i=0;i<G[p].size();i++)            {                int v = e[G[p][i]].v, w = e[G[p][i]].c;                if (dis[p]+w<dis[v])                {                    dis[v] = dis[p]+e[G[p][i]].c;                    if (!vi[v])                    {                        vi[v] = true;                        cir[v]++; if (cir[v]>n) return -1;                        q.push(v);                    }                }            }        }        return dis[to];    }}sd;int n, d, kase = 0;void init(){    scanf("%d%d", &n, &d);    sd.init(n);    for (int i=1;i<=n;i++)    {        scanf("%d", &h[i].hi);        h[i].no = i;    }    sort(h+1, h+1+n);}void solve(){    for (int i=1;i<n;i++)    {        sd.addedge(i, i+1, -1);        int u = min(h[i].no, h[i+1].no);        int v = max(h[i].no, h[i+1].no);        sd.addedge(v, u, d);    }    int u = min(h[1].no, h[n].no), v = max(h[1].no, h[n].no);    printf("Case %d: %d\n", ++kase, sd.solve(u, v));}int main(){    #ifndef ONLINE_JUDGE    freopen(FN2".in","r",stdin);    freopen(FN2".out","w",stdout);    #endif    int t; scanf("%d", &t);    while (t--)    {        init();        solve();    }    return 0;}

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