Bzoj 3144 Hnoi 2013 cut Cake min cut

The main idea: to give a three-dimensional lattice, not a point is likely to be cut, the price is the weight of this point. The height difference of the adjacent cutting points cannot exceed D, and the minimum cost makes the top and bottom separate.

Idea: Very bare minimum cut model, very God-built map.

S-> the first floor, F:inf.

All points--the dots below it, f:inf

A point of entry-and-out of a point, f:val[i]

(i,j,k)-> (i-d,j,k), F:inf

The bottom layer of the dot->t:f:inf

Then the smallest cut is the answer.

Why see: Http://www.cnblogs.com/zyfzyf/p/4182168.html Otz ZYF

CODE:

#include <queue> #include <cstdio> #include <cstring> #include <iostream> #include < algorithm> #define MAX 130000#define maxe 1200000#define S 0#define T (MAX-1) #define INF 0x3f3f3f3fusing namespace std ; const int dx[] = {0,1,-1,0,0};const int dy[] = {0,0,0,1,-1};struct maxflow{int head[max],total;int next[maxe],aim[maxe], Flow[maxe];int Deep[max]; Maxflow () {total = 1;memset (head,0,sizeof (Head));} void Add (int x,int y,int f) {next[++total] = head[x];aim[total] = Y;flow[total] = f;head[x] = total;} void Insert (int x,int y,int f) {Add (x,y,f); ADD (y,x,0);} BOOL BFS () {static queue<int> Q;while (!q.empty ()) Q.pop () memset (deep,0,sizeof (deep));d eep[s] = 1;q.push (S); while (!q.empty ()) {int x = Q.front (), Q.pop (), for (int i = head[x]; i; i = Next[i]) if (Flow[i] &&!deep[aim[i]) {de Ep[aim[i]] = deep[x] + 1;q.push (Aim[i]), if (aim[i] = = T) return true;}} return false;} int dinic (int x,int f) {if (x = = T) return F;int temp = f;for (int i = head[x]; i; i = Next[i]) if (flow[I] && temp && deep[aim[i]] = = Deep[x] + 1) {int away = Dinic (Aim[i],min (flow[i],temp)); if (!away) deep[aim[i ]] = 0;flow[i]-= away;flow[i^1] + = away;temp-= away;} return f-temp;}} Solver;int p,q,r,d;int Src[50][50][50],num[50][50][50],cnt;int Main () {cin >> P >> Q >> R >> d;for (int i = 1; I <= R; ++i) for (int j = 1; j <= P; ++j) for (int k = 1; k <= Q; ++k) {scanf ("%d", &src[i][j][k]), num I [j] [K] = ++cnt;solver. Insert (Num[i][j][k] << 1,num[i][j][k] << 1|1,src[i][j][k]);} for (int i = 1, i <= P; ++i) for (int j = 1; j <= Q; ++j) {solver. Insert (S,num[1][i][j] << 1,inf); Solver. Insert (Num[r][i][j] << 1|1,t,inf);} for (int i = 1, i <= R; ++i) for (int j = 1, J <= P; ++j) for (int k = 1; k <= Q; ++k) {if (i! = R) solver. Insert (Num[i][j][k] << 1|1,num[i + 1][j][k] << 1,inf); for (int d = 1; d <= 4; ++d) {int FX = j + dx[d],fy = K + dy[d];if (!fx | |!fy | | FX > P | | fy > Q) continue;if (i-d > 0) solvEr. Insert (Num[i][j][k] << 1,num[i-d][fx][fy] << 1,inf);}} int max_flow = 0;while (solver. BFS ()) Max_flow + = Solver. Dinic (S,inf); cout << Max_flow << endl;return 0;}

Bzoj 3144 Hnoi 2013 cut Cake min cut