题目大意
给出一个网格图，每个格子上有移动次数限制。每次可以交换相邻的两个棋子（有公共点就算相邻）。给出一个初始状态，问最少需要多少步达到目标状态。
思路
这个题主要是限制是每个格子，而不是棋子。我们对每个格子拆点，，相邻的格子之间连边，经过一个格子的时候的费用是2，流量是（正常的流量+这个点是入点+这个点是出点）/2，在连S和T的时候要将费用设成-1。这样跑出来的最小费用的一半就是答案。 注意要特判一下起始情况和目标情况黑子不同的情况。。
CODE;const int dx[9] = {0, -1, -1, -1, 0, 0, 1, 1, 1};const int dy[9] = {0, -1, 0, 1, -1, 1, -1, 0, 1};int blacks, _blacks;struct MinCostMaxFlow{int head[MAXP], total;int _next[MAXE], aim[MAXE], flow[MAXE], cost[MAXE];int f[MAXP], from[MAXP], p[MAXP];bool v[MAXP];MinCostMaxFlow():total(1) {}void Add(int x, int y, int f, int c) {_next[++total] = head[x];aim[total] = y;flow[total] = f;cost[total] = c;head[x] = total;}void Insert(int x, int y, int f, int c) {Add(x, y, f, c);Add(y, x, 0, -c);}bool SPFA() {static queue<int> q;while(!q.empty()) q.pop();memset(f, 0x3f, sizeof(f));memset(v, false, sizeof(v));f[S] = 0;q.push(S);while(!q.empty()) {int x = q.front(); q.pop();v[x] = false;for(int i = head[x]; i; i = _next[i])if(flow[i] && f[aim[i]] > f[x] + cost[i]) {f[aim[i]] = f[x] + cost[i];if(!v[aim[i]])v[aim[i]] = true, q.push(aim[i]);from[aim[i]] = x;p[aim[i]] = i;}}return f[T] != INF;}int EdmondsKarp() {int re = 0, total_flow = 0;while(SPFA()) {int max_flow = INF;for(int i = T; i != S; i = from[i])max_flow = min(max_flow, flow[p[i]]);for(int i = T; i != S; i = from[i]) {flow[p[i]] -= max_flow;flow[p[i] ^ 1] += max_flow;}re += f[T] * max_flow;total_flow += max_flow;}return total_flow == blacks ? (re >> 1) : -1;}}solver;int m, n;char s[MAX][MAX];bool _in[MAX][MAX], _out[MAX][MAX];int src[MAX][MAX];int num[MAX][MAX], cnt;int main(){cin >> m >> n;for(int i = 1; i <= m; ++i) {scanf(“%s”, s[i] + 1);for(int j = 1; j <= n; ++j) {_in[i][j] = s[i][j] == ‘1’;blacks += _in[i][j];}}for(int i = 1; i <= m; ++i) {scanf(“%s”, s[i] + 1);for(int j = 1; j <= n; ++j) {_out[i][j] = s[i][j] == ‘1’;_blacks += _out[i][j];}}if(blacks != _blacks) {puts(“-1”);return 0;}for(int i = 1; i <= m; ++i) {scanf(“%s”, s[i] + 1);for(int j = 1; j <= n; ++j)src[i][j] = s[i][j] – ‘0’;}for(int i = 1; i <= m; ++i)for(int j = 1; j <= n; ++j)num[i][j] = ++cnt;for(int i = 1; i <= m; ++i)for(int j = 1; j <= n; ++j) {if(_in[i][j])solver.Insert(S, num[i][j] << 1, 1, -1);if(_out[i][j])solver.Insert(num[i][j] << 1|1, T, 1, -1);solver.Insert(num[i][j] << 1, num[i][j] << 1|1, (src[i][j] + _in[i][j] + _out[i][j]) >> 1, 2);for(int k = 1; k <= 8; ++k) {int fx = i + dx[k], fy = j + dy[k];if(!fx || !fy || fx > m || fy > n) continue;solver.Insert(num[i][j] << 1|1, num[fx][fy] << 1, INF, 0);}}cout << solver.EdmondsKarp() << endl;return 0;}
你在无垠的海边第一次听到了自己心跳的声音，
