这题可以说是最小割模型最容易的一道题目了。根据割的定义我们知道，割其实就是将原图的点集分成两个部分(s',t')。两个部分分别包含源点S，汇点T。那这道题目里，就是要把点集分成羊一个集合(sheep)，狼一个集合(wolf)。S->sheepwolf->T因为在矩阵中，可以随便走。所以矩阵中相邻的两点建立双向边，容量为1。求最小割，就是答案
#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#define DIR2#define MAXVERTEX 40001#define MAXARC160004#define INF0x7ffffffusing namespace std;struct Edge{int v, val, next;}edge[MAXARC];int e_cnt, head[MAXVERTEX];int pre[MAXVERTEX], cur[MAXVERTEX], dis[MAXVERTEX], gap[MAXVERTEX];int dir[][2] = { {1, 0}, {0, 1} };void insert_arc(int u, int v, int w1){edge[e_cnt].val = w1;edge[e_cnt].v = v;edge[e_cnt].next = head[u];head[u] = e_cnt ++;swap(u, v);edge[e_cnt].val = 0;edge[e_cnt].v = v;edge[e_cnt].next = head[u];head[u] = e_cnt ++;}void bfs(int src, int des){int u, v;queue<int> q;memset(dis, -1, sizeof(dis));memset(gap, 0, sizeof(gap));dis[des] = 0;gap[0] = 1;q.push(des);while( !q.empty() ) {u = q.front(); q.pop();for(int i = head[u]; -1 != i; i = edge[i].next) {v = edge[i].v;if( edge[i].val || -1 != dis[v] )continue;dis[v] = dis[u]+1;gap[dis[v]] ++;q.push(v);}}}int i_sap(int source, int sink, int e_v){int u(source), v, min_dis, max_flow(0), path_flow(INF);bfs(source, sink);for(int i = 0; i <= e_v; i ++)cur[i] = head[i];pre[u] = u;while( dis[source] < e_v ) {loop:for(int &i = cur[u]; -1 != i; i = edge[i].next) {v = edge[i].v;if( !edge[i].val || (dis[u] != dis[v]+1) )continue;pre[v] = u;path_flow = min(path_flow, edge[i].val);u = v;if( v == sink ) {for(u = pre[u]; v != source; v = u, u = pre[u]) {edge[cur[u]].val -= path_flow;edge[cur[u]^1].val += path_flow;}max_flow += path_flow;path_flow = INF;}goto loop;}min_dis = e_v;for(int i = head[u]; -1 != i; i = edge[i].next) {v = edge[i].v;if( edge[i].val && min_dis > dis[v] ) {min_dis = dis[v]; cur[u] = i;}}gap[dis[u]] --;if( !gap[dis[u]] )break;dis[u] = min_dis+1;gap[ dis[u] ] ++;u = pre[u];}//printf("max_flow = %d\n", max_flow);return max_flow;}void debug(int e){printf("e = %d\n", e);int r[100][100];memset(r, 0, sizeof(r));for(int u = 0; u <= e; u ++) {for(int v = head[u]; v != -1; v = edge[v].next ){ r[u][edge[v].v] = edge[v].val;}}for(int u = 0; u <= e; u ++) {for(int v = 0; v <= e; v ++) {if( r[u][v] == INF ) printf("X ");elseprintf("%d ", r[u][v]);}printf("\n");}printf("\n");}int main(int argc, char const *argv[]){//freopen("test.in", "r", stdin);int row, col, vs, x, y, src, des, cnt(1), pos;while( ~scanf("%d %d", &row, &col) ) {e_cnt = 0;src = row*col;des = src+1;memset(head, -1, sizeof(head));int c = 0;for(int i = 0; i < row; i ++) {for(int j = 0; j < col; j ++) {scanf("%d", &vs); pos = i*col+j;for(int d = 0; d < DIR; d ++) {x = i+dir[d][0]; y = j+dir[d][1];if( x < 0 || y < 0 || x >= row || y >= col )continue;insert_arc(pos, x*col+y, 1);insert_arc(x*col+y, pos, 1);}if( 1 == vs )insert_arc(src, pos, INF);else if( 2 == vs )insert_arc(pos, des, INF);}}printf("Case %d:\n%d\n", cnt ++, i_sap(src, des, des));}return 0;}
