费用流 hdu3376 Matrix Again
传送门:点击打开链接
n太大了所以不能使用O(n^4)的dp来做了,仅仅能考虑更复杂的费用流
基本的问题还是难在怎样建图
将点拆分成两个,分别用i和i+n*n来表示
对于n*n个点,从i到i+n*n建一条边费用是-A[i][j]容量是1。来表示路过(i,j)获取的值
然后从第二层建两条边。连向右边和下边的两个点的第一层,让点再次回到第一层。费用是0
这样下来。点从第一层到第二层,表示获取了这个点的值,并且假设不获取这个值,就无法到达第二层,就无法继续扩展了。
所以就限制了每一个点仅仅能走一次了~
最后在考虑怎样让(1,1)和(n,n)的费用仅仅算一次,从源点出来后一个连向(1,1)的第一层。一个连向(1,1)的第二层。这样(1,1)的值就仅仅会被算一次
汇点也是一样。从(n,n)的第一层和(n,n)的第二层都连向汇点即可了
然后再加上读入挂,。用G++提交,就不会TLE了
#include<map> #include<set> #include<cmath> #include<stack> #include<queue> #include<cctype> #include<cstdio> #include<string> #include<vector> #include<cstring> #include<iostream> #include<algorithm> #include<functional> #define FIN freopen("input.txt","r",stdin) #define FOUT freopen("output.txt","w+",stdout) using namespace std; typedef long long LL; typedef pair<int, int> PII; const int MX = 1e6 + 5;//都开4倍把.. const int MM = 3e6 + 5; const int MS = 1000 + 5; const int INF = 0x3f3f3f3f; struct Edge { int to, next, cap, flow, cost; Edge() {} Edge(int _to, int _next, int _cap, int _flow, int _cost) { to = _to; next = _next; cap = _cap; flow = _flow; cost = _cost; } } E[MM]; int Head[MX], tol; int pre[MX]; //储存前驱顶点 int dis[MX]; //储存到源点s的距离 bool vis[MX]; int N;//节点总个数,节点编号从0~N-1 void init(int n) { tol = 0; N = n + 2; memset(Head, -1, sizeof(Head)); } void edge_add(int u, int v, int cap, int cost) { E[tol] = Edge(v, Head[u], cap, 0, cost); Head[u] = tol++; E[tol] = Edge(u, Head[v], 0, 0, -cost); Head[v] = tol++; } bool spfa(int s, int t) { queue<int>q; for (int i = 0; i < N; i++) { dis[i] = INF; vis[i] = false; pre[i] = -1; } dis[s] = 0; vis[s] = true; q.push(s); while (!q.empty()) { int u = q.front(); q.pop(); vis[u] = false; for (int i = Head[u]; i != -1; i = E[i].next) { int v = E[i].to; if (E[i].cap > E[i].flow && dis[v] > dis[u] + E[i].cost) { dis[v] = dis[u] + E[i].cost; pre[v] = i; if (!vis[v]) { vis[v] = true; q.push(v); } } } } if (pre[t] == -1) return false; else return true; } //返回的是最大流, cost存的是最小费用 int minCostMaxflow(int s, int t, int &cost) { int flow = 0; cost = 0; while (spfa(s, t)) { int Min = INF; for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) { if (Min > E[i].cap - E[i].flow) Min = E[i].cap - E[i].flow; } for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) { E[i].flow += Min; E[i ^ 1].flow -= Min; cost += E[i].cost * Min; } flow += Min; } return flow; } int n, A[MS][MS]; int dist[][2] = {{0, 1}, {1, 0}}; inline int ID(int x, int y, int p = 0) { return (x - 1) * n + y + p * n * n; } inline int read() { char c = getchar(); while(!isdigit(c)) c = getchar(); int x = 0; while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); } return x; } int main() { //FIN; while(~scanf("%d", &n)) { int s = 0, t = 2 * n * n; init(t); edge_add(s, ID(1, 1), 1, 0); edge_add(s, ID(1, 1, 1), 1, 0); edge_add(ID(n, n), t, 1, 0); edge_add(ID(n, n, 1), t, 1, 0); for(int i = 1; i <= n; i++) { for(int j = 1; j <= n; j++) { A[i][j] = read(); edge_add(ID(i, j), ID(i, j, 1), 1, -A[i][j]); for(int k = 0; k < 2; k++) { int nx = i + dist[k][0]; int ny = j + dist[k][1]; if(nx <= n && ny <= n) { edge_add(ID(i, j, 1), ID(nx, ny), 1, 0); } } } } int ans = 0; minCostMaxflow(s, t, ans); printf("%d\n", -ans); } return 0; }