【費用流】HDU 5406 CRB and Apple

標籤：

通道

題意：

思路：

代碼：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;typedef long long ll;template <class T>inline bool rd(T &ret) {    char c; int sgn;    if(c = getchar() , c == EOF) return false;    while(c != ‘-‘ && (c < ‘0‘ || c > ‘9‘)) c = getchar();    sgn = (c == ‘-‘) ? -1 : 1;    ret = (c == ‘-‘) ? 0 : (c - ‘0‘);    while(c = getchar(), c >= ‘0‘ && c <= ‘9‘) ret = ret * 10 + (c - ‘0‘);    ret *= sgn;    return true;}const int inf = 0x3f3f3f3f;const int MAX_N = 2007;const int MAX_M = MAX_N * MAX_N;struct Edge {    int u, v, cap, cost, nxt;     Edge () {            }    Edge (int _u, int _v, int _C, int _c, int _n) {        u = _u, v = _v, cap = _C, cost = _c, nxt = _n;    }};struct edge {    int u, v, cap, flow, cost, nxt;    void set(int _u, int _v, int _cap, int _flow, int _cost, int _nxt) {        u = _u, v = _v, cap = _cap, flow = _flow, cost = _cost, nxt = _nxt;    }};struct mcmf {    int fst[MAX_N], cc, d[MAX_N], p[MAX_N], a[MAX_N];    edge e[MAX_M];    bool in[MAX_N];    void init() {        memset(fst, -1, sizeof(fst)); cc = 0;    }    void add(int u, int v, int cap, int cost) {        e[cc].set(u, v, cap, 0, cost, fst[u]), fst[u] = cc++;        e[cc].set(v, u, 0, 0, -cost, fst[v]), fst[v] = cc++;    }    int spfa(int s, int t, int &mf, int &mc) {        memset(d, 0x3f, sizeof(d));        memset(in, 0, sizeof(in));        d[s] = 0, a[s] = inf, in[s] = 1, p[s] = 0;        queue<int> q; q.push(s);        while(!q.empty()) {            int u = q.front(); q.pop(); in[u] = 0;            for(int i = fst[u]; ~i; i = e[i].nxt) {                int v = e[i].v;                if(e[i].cap > e[i].flow && d[v] > d[u] + e[i].cost) {                    d[v] = d[u] + e[i].cost, p[v] = i;                    a[v] = min(a[u], e[i].cap - e[i].flow);                    if(!in[v]) in[v] = 1, q.push(v);                }            }        }        if(d[t] == inf) return 0;        mf += a[t], mc += a[t] * d[t];        int u = t;        while(u != s) {            e[p[u]].flow += a[t], e[p[u] ^ 1].flow -= a[t];            u = e[p[u]].u;        }        return 1;    }    int go(int s, int t) {        int ret = 0, mf = 0, mc = 0;        while(spfa(s, t, mf, mc)) ret = min(ret, mc);        return ret;    }}go;struct Node {    int H, D;    Node () {            }    Node (int _H, int _D) {        H = _H, D = _D;    }    bool operator < (const Node &rhs) const {        if (H != rhs.H) return H > rhs.H;        return D < rhs.D;    }} a[MAX_N];int m;struct Bit {    ll bit[MAX_N];    void clear() {        memset(bit, 0, sizeof bit);    }    void add(int i, ll v) {        if (i == 0) bit[0] += v;        else {            for (;i <= m; i += i & -i)                bit[i] += v;        }    }    ll query(int i) {        if (i < 0) return 0;        else {            ll re = 0;            for (;i > 0; i -= i & -i)                re += bit[i];            return re;        }    }};int n, b[MAX_N], cnt[MAX_N][MAX_N];Bit Tree;void Clear() {    Tree.clear();    go.init();}int main() {    int T; rd(T);    while (T-- > 0) {        Clear();        rd(n);        m = 0;        for (int i = 0; i < n; ++i) {            rd(a[i].H), rd(a[i].D);            b[++m] = a[i].D;        }        if (n <= 2) {            printf("%d\n", n);            continue;        }        sort(b + 1, b + 1 + m);        m = unique(b + 1, b + n + 1) - b - 1;        sort(a, a + n);        for (int i = 0; i < n; ++i)             a[i].D = lower_bound(b + 1, b + m + 1, a[i].D) - b;        for (int i = 0; i < n; ++i) {            for (int j = 1; j <= m; ++j) Tree.bit[j] = 0;            for (int j = i; j < n; ++j) {                if (a[j].D < a[i].D) continue;                Tree.add(a[j].D, 1);                cnt[i][j] = Tree.query(a[j].D);            }        }        int S = 0, T = 2 * n + 1, X = T + 1;        go.add(S, X, 2, 0);        for (int i = 0; i < n; ++i) {            go.add(X, i * 2 + 1, 1, 0);            go.add(i * 2 + 2, T, 1, 0);                go.add(i * 2 + 1, i * 2 + 2, 1, 0);            for (int j = i + 1; j < n; ++j) {                if (a[i].D > a[j].D) continue;                if (cnt[i][j] >= 5) continue;                go.add(i * 2 + 2, j * 2 + 1, 1, -1);            }        }        int ans = -go.go(S, T) + 2;        printf("%d\n", ans);    }    return 0;}

View Code

 

【費用流】HDU 5406 CRB and Apple