[SWUST1759] 骑士共存问题（最大流，最大独立集）

题目链接：https://www.oj.swust.edu.cn/problem/show/1759

和方格取数一样建模就行了。

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3
  4 typedef struct Edge {
  5     int u, v, w, next;
  6 }Edge;
  7
  8 const int inf = 0x7f7f7f7f;
  9 const int maxn = 44444;
 10
 11 int cnt, dhead[maxn];
 12 int cur[maxn], dd[maxn];
 13 Edge dedge[800001];
 14 int S, T, N;
 15
 16 void init() {
 17     memset(dhead, -1, sizeof(dhead));
 18     for(int i = 0; i < maxn; i++) dedge[i].next = -1;
 19     S = 0; cnt = 0;
 20 }
 21
 22 void adde(int u, int v, int w, int c1=0) {
 23     dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w;
 24     dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
 25     dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1;
 26     dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
 27 }
 28
 29 bool bfs(int s, int t, int n) {
 30     queue<int> q;
 31     for(int i = 0; i < n; i++) dd[i] = inf;
 32     dd[s] = 0;
 33     q.push(s);
 34     while(!q.empty()) {
 35         int u = q.front(); q.pop();
 36         for(int i = dhead[u]; ~i; i = dedge[i].next) {
 37             if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) {
 38                 dd[dedge[i].v] = dd[u] + 1;
 39                 if(dedge[i].v == t) return 1;
 40                 q.push(dedge[i].v);
 41             }
 42         }
 43     }
 44     return 0;
 45 }
 46
 47 int dinic(int s, int t, int n) {
 48     int st[maxn], top;
 49     int u;
 50     int flow = 0;
 51     while(bfs(s, t, n)) {
 52         for(int i = 0; i < n; i++) cur[i] = dhead[i];
 53         u = s; top = 0;
 54         while(cur[s] != -1) {
 55             if(u == t) {
 56                 int tp = inf;
 57                 for(int i = top - 1; i >= 0; i--) {
 58                     tp = min(tp, dedge[st[i]].w);
 59                 }
 60                 flow += tp;
 61                 for(int i = top - 1; i >= 0; i--) {
 62                     dedge[st[i]].w -= tp;
 63                     dedge[st[i] ^ 1].w += tp;
 64                     if(dedge[st[i]].w == 0) top = i;
 65                 }
 66                 u = dedge[st[top]].u;
 67             }
 68             else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
 69                 st[top++] = cur[u];
 70                 u = dedge[cur[u]].v;
 71             }
 72             else {
 73                 while(u != s && cur[u] == -1) {
 74                     u = dedge[st[--top]].u;
 75                 }
 76                 cur[u] = dedge[cur[u]].next;
 77             }
 78         }
 79     }
 80     return flow;
 81 }
 82
 83 typedef pair<int, int> pii;
 84 const int dx[11] = {-1,1,2,2,1,-1,-2,-2};
 85 const int dy[11] = {-2,-2,-1,1,2,2,1,-1};
 86 const int maxm = 1010;
 87 int n, m;
 88 int G[maxm][maxm], icnt;
 89 map<pii, int> shawanyiera;
 90
 91 inline int id(int x, int y) {
 92     return shawanyiera[pii(x, y)];
 93 }
 94
 95 inline bool ok(int x, int y) {
 96         return x >= 1 && x <= n && y >= 1 && y <= n;
 97 }
 98
 99 int main() {
100     // freopen("in", "r", stdin);
101     int u, v;
102     scanf("%d%d",&n,&m);
103     memset(G, 0, sizeof(G));
104     init(); shawanyiera.clear(); icnt = 0;
105     for(int i = 0; i < m; i++) {
106         scanf("%d%d",&u,&v);
107         G[u][v] = 1;
108     }
109     for(int i = 1; i <= n; i++) {
110         for(int j = 1; j <= n; j++) {
111             if(G[i][j]) continue;
112             shawanyiera[pii(i, j)] = ++icnt;
113         }
114     }
115     S = 0, T = 2 * icnt + 1, N = T + 1;
116     for(int i = 1; i <= icnt; i++) {
117         adde(S, i, 1);
118         adde(i+icnt, T, 1);
119     }
120     for(int i = 1; i <= n; i++) {
121         for(int j = 1; j <= n; j++) {
122             if(G[i][j]) continue;
123             for(int k = 0; k < 8; k++) {
124                 int x = i + dx[k];
125                 int y = j + dy[k];
126                 if(!ok(x, y)) continue;
127                 if(G[x][y]) continue;
128                 adde(id(i,j), icnt+id(x,y), inf);
129             }
130         }
131     }
132     // cout << icnt << endl;
133     printf("%d\n", icnt-dinic(S, T, N)/2);
134
135     return 0;
136 }