一开始还以为是个水题... 没想到是个藏于市井之中的dalao
正常来讲想的肯定是贪心或者dp
但是今天练的是网络流同时有行和列的限制或者中间空一行什么的
然后就正难则反 求一下舍弃的点的最小值
所以按邻接性染色 横纵坐标和为奇数的连源点 反之连汇点
然后相邻点之间连一条inf表示两个必须切一个
然后跑最小割就行
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<iostream> 5 #include<algorithm> 6 #define ms(a,b) memset(a,b,sizeof a) 7 #define rep(i,a,n) for(int i = a;i <= n;i++) 8 #define per(i,n,a) for(int i = n;i >= a;i--) 9 #define inf 2147483647 10 using namespace std; 11 typedef long long ll; 12 ll read() { 13 ll as = 0,fu = 1; 14 char c = getchar(); 15 while(c < '0' || c > '9') { 16 if(c == '-') fu = -1; 17 c = getchar(); 18 } 19 while(c >= '0' && c <= '9') { 20 as = as * 10 + c - '0'; 21 c = getchar(); 22 } 23 return as * fu; 24 } 25 const int N = 10005; 26 const int M = 200005; 27 //head 28 int s = N-2,t = N-1; 29 int head[N],nxt[M],mo[M],cnt = 1; 30 ll cst[M]; 31 void _add(int x,int y,ll w) { 32 mo[++cnt] = y; 33 cst[cnt] = w; 34 nxt[cnt] = head[x]; 35 head[x] = cnt; 36 } 37 void add(int x,int y,ll w) { 38 if(x^y) _add(x,y,w),_add(y,x,0); 39 } 40 41 int dep[N],cur[N]; 42 bool bfs() { 43 queue<int> q; 44 memcpy(cur,head,sizeof cur); 45 ms(dep,0),q.push(s),dep[s] = 1; 46 while(!q.empty()) { 47 int x = q.front(); 48 q.pop(); 49 for(int i = head[x];i;i = nxt[i]) { 50 int sn = mo[i]; 51 if(!dep[sn] && cst[i]) { 52 dep[sn] = dep[x] + 1; 53 q.push(sn); 54 } 55 } 56 } 57 return dep[t]; 58 } 59 60 ll dfs(int x,ll flow) { 61 if(x == t || flow == 0ll) return flow; 62 ll res = 0; 63 for(int &i = cur[x];i;i = nxt[i]) { 64 int sn = mo[i]; 65 if(dep[sn] == dep[x] + 1 && cst[i]) { 66 ll d = dfs(sn,min(cst[i],flow - res)); 67 if(d) { 68 cst[i] -= d,cst[i^1] += d; 69 res += d; 70 if(res == flow) break; 71 } 72 } 73 } 74 if(res ^ flow) dep[x] = 0; 75 return res; 76 } 77 78 ll DINIC() { 79 ll ans = 0; 80 while(bfs()) ans += dfs(s,inf); 81 return ans; 82 } 83 84 int n,m; 85 int idx(int x,int y) {return (x-1)*m+y;} 86 87 int dx[] = {0,1,0,-1}; 88 int dy[] = {1,0,-1,0}; 89 bool check(int x,int y,int d) { 90 x += dx[d],y += dy[d]; 91 if(x <= 0 || y <= 0 || x > n || y > m) return 0; 92 return 1; 93 } 94 ll a[105][105],sum; 95 96 int main() { 97 n = read(),m = read(); 98 rep(i,1,n) rep(j,1,m) { 99 int x = read(); 100 sum += x; 101 if(i+j & 1) add(s,idx(i,j),x); 102 else add(idx(i,j),t,x); 103 } 104 rep(i,1,n) rep(j,1,m) { 105 if(!(i+j & 1)) continue; 106 rep(d,0,3) if(check(i,j,d)) 107 add(idx(i,j),idx(i+dx[d],j+dy[d]),inf); 108 } 109 sum -= DINIC(); 110 printf("%lld ",sum); 111 return 0; 112 }
附luogu大样例一组
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