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  • bzoj2132: 圈地计划

    要分成两坨对吧。。 所以显然最小割

    但是不兹辞啊。。

    最小割是最小的啊 求最大费用怎么玩啊

    那咱们就把所有费用都加起来,减掉一个最小的呗

    但是两个属于不同集合的点贡献的价值是负的啊

    网络流怎么跑负的啊

    那咱就交换一下呗

    原图是二分图啊,把另一部分与S和T连边的流量换一下就好啦。

    注意哦 n和m可能为1 所以累加C的时候写成ans += C[i][j] * (4 - (i == 1 || i == n) - (j == 1 || j == m));就错啦。

    要么在加边的时候累加,要么写成ans += C[i][j] * ((i != 1) + (i != n) + (j != 1) + (j != m)); 

      1 #include<cstdio>
      2 #include<cstring>
      3 #include<cstdlib>
      4 #include<algorithm>
      5 #include<iostream>
      6 #define rep(i, a, b) for(int i = (a), _end = (b); i <= _end; ++i)
      7 
      8 using namespace std;
      9 
     10 void setIO(const string& s) {
     11     freopen((s + ".in").c_str(), "r", stdin);
     12     freopen((s + ".out").c_str(), "w", stdout);
     13 }
     14 template<typename Q> Q read(Q& x) {
     15     static char c, f;
     16     for(f = 0; c = getchar(), !isdigit(c); ) if(c == '-') f = 1;
     17     for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - '0';
     18     if(f) x = -x;
     19     return x;
     20 }
     21 template<typename Q> Q read() {
     22     static Q x; read(x); return x;
     23 }
     24 
     25 // ISAP
     26 const int N = 100 * 100 + 10, M = 100 * 100 * 4 + 10, INF = ~0u >> 1;
     27 
     28 struct Edge {
     29     int to, adv;
     30     Edge *next;
     31     Edge(int to = 0, int adv = 0, Edge *next = 0) : to(to), adv(adv), next(next) {}
     32 }pool[M * 2], *pis = pool, *fir[N], *pre[N];
     33 
     34 void AddEdge(int u, int v, int w, int b = 0) {
     35     fir[u] = new(pis++) Edge(v, w, fir[u]);
     36     fir[v] = new(pis++) Edge(u, w * b, fir[v]);
     37 }
     38 
     39 Edge *inv(Edge *p) {
     40     return pool + ((p - pool) ^ 1);
     41 }
     42 
     43 int d[N], q[N], ql, qr;
     44 
     45 namespace ISAP {
     46     int n, s, t;
     47     int num[N];
     48     Edge *cur[N];
     49     
     50     void insert(int u, int dis) {
     51         if(d[u] == n) {
     52             d[u] = dis;
     53             q[qr++] = u;
     54         }
     55     }
     56     
     57     void BFS() {
     58         for(int u = 1; u <= n; u++) {
     59             d[u] = n;
     60         }
     61         ql = qr = 0;
     62         insert(t, 0);
     63         while(ql ^ qr) {
     64             int u = q[ql++];
     65             for(Edge *p = fir[u]; p; p = p->next) if(inv(p)->adv){
     66                 insert(p->to, d[u] + 1);
     67             }
     68         }
     69     }
     70     
     71     int Augment() {
     72         int a = INF;
     73         for(int u = t; u != s; u = inv(pre[u])->to) {
     74             a = min(a, pre[u]->adv);
     75         }
     76         for(int u = t; u != s; u = inv(pre[u])->to) {
     77             pre[u]->adv -= a;
     78             inv(pre[u])->adv += a;
     79         }
     80         return a;
     81     }
     82     
     83     int Maxflow() {
     84         memset(num, 0, sizeof num);
     85         BFS();
     86         for(int u = 1; u <= n; u++) {
     87             cur[u] = fir[u];
     88             num[d[u]]++;
     89         }
     90         for(int i = 1; i < d[s]; i++) {
     91             if(!num[i]) return 0;
     92         }
     93         int flow = 0;
     94         for(int u = s; d[s] < n; ) {
     95             if(u == t) {
     96                 flow += Augment();
     97                 u = s;
     98             }
     99             int ok = 0;
    100             for(Edge *&p = cur[u]; p; p = p->next) if(p->adv) {
    101                 int v = p->to;
    102                 if(d[v] + 1 == d[u]) {
    103                     pre[u = v] = p;
    104                     ok = 1;
    105                     break;
    106                 }
    107             }
    108             if(!ok) {
    109                 int &dis = d[u];
    110                 if(!--num[dis]) break;
    111                 dis = n;
    112                 for(Edge *p = (cur[u] = fir[u]); p; p = p->next) if(p->adv) {
    113                     dis = min(dis, d[p->to] + 1);
    114                 }
    115                 num[dis]++;
    116                 if(u != s) u = inv(pre[u])->to;
    117             }
    118         }
    119         return flow;
    120     }
    121     
    122     int main(int _n, int _s, int _t) {
    123         n = _n, s = _s, t = _t;
    124         return Maxflow();
    125     }
    126 }
    127 
    128 int n, m;
    129 
    130 int encode(int x, int y) {
    131     return (x - 1) * m + y;
    132 }
    133 
    134 const int maxn = 100 + 10;
    135 
    136 int A[maxn][maxn], B[maxn][maxn], C[maxn][maxn];
    137 
    138 int main() {
    139 #ifdef DEBUG
    140     freopen("in.txt", "r", stdin);
    141     freopen("out.txt", "w", stdout);
    142 #endif
    143     
    144     read(n), read(m);
    145     int ans = 0;
    146     rep(i, 1, n) rep(j, 1, m) ans += read(A[i][j]);
    147     rep(i, 1, n) rep(j, 1, m) ans += read(B[i][j]);
    148     rep(i, 1, n) rep(j, 1, m) read(C[i][j]);
    149     int s = n * m + 1, t = s + 1;
    150     rep(i, 1, n) rep(j, 1, m) {
    151         int u = encode(i, j);
    152         if((i ^ j) & 1) swap(A[i][j], B[i][j]);
    153         AddEdge(s, u, A[i][j]);
    154         AddEdge(u, t, B[i][j]);
    155         if(i < n) AddEdge(u, encode(i + 1, j), C[i][j] + C[i+1][j], 1), ans += C[i][j] + C[i+1][j];
    156         if(j < m) AddEdge(u, encode(i, j + 1), C[i][j] + C[i][j+1], 1), ans += C[i][j] + C[i][j+1];
    157     }
    158     
    159     printf("%d
    ", ans - ISAP::main(t, s, t));
    160     
    161     return 0;
    162 }
    View Code
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  • 原文地址:https://www.cnblogs.com/showson/p/5043565.html
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