一眼看就是最小割板子题，建图也很直观，注意每一条边建双向边其实不用建4条边，只要反向边的容量和正边相同就行。然后直接跑最大流板子就行。不过此题拿vector存图会MLE……而拿链前存图就能卡过去……场面一度十分尴尬。
这里发一个vector80分代码，各位改成链前就能AC了……
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<queue> 8 #include<stack> 9 #include<vector> 10 #include<cctype> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 1e6 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) 27 { 28 ans = ans * 10 + ch - '0'; ch = getchar(); 29 } 30 if(last == '-') ans = -ans; 31 return ans; 32 } 33 inline void write(ll x) 34 { 35 if(x < 0) putchar('-'), x = -x; 36 if(x >= 10) write(x / 10); 37 putchar(x % 10 + '0'); 38 } 39 40 int n, m; 41 42 struct Edge 43 { 44 int from, to, cap, flow; 45 }; 46 vector<Edge> edges; 47 vector<int> G[maxn]; 48 void addedge(int from, int to, int w) 49 { 50 edges.push_back((Edge){from, to, w, 0}); 51 edges.push_back((Edge){to, from, w, 0}); 52 int sz = edges.size(); 53 G[from].push_back(sz - 2); 54 G[to].push_back(sz - 1); 55 } 56 57 int dis[maxn]; 58 bool vis[maxn]; 59 bool bfs() 60 { 61 for(rg int i = 1; i <= n * m; ++i) vis[i] = 0; 62 dis[1] = 0; vis[1] = 1; 63 queue<int> q; q.push(1); 64 while(!q.empty()) 65 { 66 int now = q.front(); q.pop(); 67 for(rg int i = 0; i < (int)G[now].size(); ++i) 68 { 69 Edge& e = edges[G[now][i]]; 70 if(!vis[e.to] && e.cap > e.flow) 71 { 72 vis[e.to] = 1; 73 dis[e.to] = dis[now] + 1; 74 q.push(e.to); 75 } 76 } 77 } 78 return vis[n * m]; 79 } 80 int cur[maxn]; 81 int dfs(const int& now, int a) 82 { 83 if(now == n * m || !a) return a; 84 int flow = 0, f; 85 for(int& i = cur[now]; i < (int)G[now].size(); ++i) 86 { 87 Edge& e = edges[G[now][i]]; 88 if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) 89 { 90 e.flow += f; 91 edges[G[now][i] ^ 1].flow -= f; 92 flow += f; a -= f; 93 if(!a) break; 94 } 95 } 96 return flow; 97 } 98 99 int maxflow() 100 { 101 int flow = 0; 102 while(bfs()) 103 { 104 for(rg int i = 1; i <= n * m; ++i) cur[i] = 0; 105 flow += dfs(1, INF); 106 } 107 return flow; 108 } 109 110 int main() 111 { 112 n = read(); m = read(); 113 for(rg int i = 1; i <= n; ++i) 114 for(rg int j = 1; j < m; ++j) 115 { 116 int x = read(), from = (i - 1) * m + j, to = (i - 1) * m + j + 1; 117 addedge(from, to, x); 118 } 119 for(rg int i = 1; i < n; ++i) 120 for(rg int j = 1; j <= m; ++j) 121 { 122 int x = read(), from = (i - 1) * m + j, to = i * m + j; 123 addedge(from, to, x); 124 } 125 for(rg int i = 1; i < n; ++i) 126 for(rg int j = 1; j < m; ++j) 127 { 128 int x = read(), from = (i - 1) * m + j, to = i * m + j + 1; 129 addedge(from, to, x); 130 } 131 write(maxflow()); enter; 132 return 0; 133 }
然后此题据说也可以用对偶图的知识解决。