題目鏈接
最大流裸題，沒什么好說吧，恰好點數多，考驗網絡流的效率，正好練\(Dinic\)。

#include <cstdio> #include <queue> #include <cstring> #define INF 2147483647 using namespace std; const int MAXN = 1000010; const int MAXM = 8000010; inline int read(){int s = 0, w = 1;char ch = getchar();while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }return s * w; } struct Edge{int next, to, rest; }e[MAXM]; int s, t, num = 1, n, m; int head[MAXN]; inline void Add(int from, int to, int flow){e[++num] = (Edge){ head[from], to, flow }; head[from] = num;e[++num] = (Edge){ head[to], from, flow }; head[to] = num; } int level[MAXN], now, sum; queue <int> q; int re(){memset(level, 0, sizeof level);while(q.size()) q.pop();q.push(s); level[s] = 1;while(q.size()){now = q.front(); q.pop();for(int i = head[now]; i; i = e[i].next)if(e[i].rest && !level[e[i].to]){level[e[i].to] = level[now] + 1;q.push(e[i].to);}}return level[t]; } int findflow(int u, int flow){if(!flow || u == t) return flow;int f = 0, t;for(int i = head[u]; i; i = e[i].next){if(e[i].rest && level[e[i].to] == level[u] + 1){f += (t = findflow(e[i].to, min(flow - f, e[i].rest)));e[i].rest -= t; e[i ^ 1].rest += t;}}if(!f) level[u] = 0;return f; } int dinic(){int ans = 0;while(re())ans += findflow(s, INF);return ans; } inline int id(int i, int j){return (i - 1) * m + j; } int main(){n = read(); m = read(); s = id(1, 1); t = id(n, m);for(int i = 1; i <= n; ++i)for(int j = 1; j < m; ++j)Add(id(i, j), id(i, j + 1), read());for(int i = 1; i < n; ++i)for(int j = 1; j <= m; ++j)Add(id(i, j), id(i + 1, j), read());for(int i = 1; i < n; ++i)for(int j = 1; j < m; ++j)Add(id(i, j), id(i + 1, j + 1), read());printf("%d\n", dinic());return 0; }

轉載于:https://www.cnblogs.com/Qihoo360/p/10326350.html

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