题目链接：点击打开链接
题意：
给定n*m的地图 a个男人 b个女人
下面n*m的方格图.表示空地 *表示障碍。
下面第一行给出中性人的坐标和移动时间
下面a行给出每个男人的坐标和移动时间
下面b行给出女人的坐标和移动时间。
移动时间是指人移动到相邻矩阵的时间（人是不能走到障碍物上的）
每个空地上恰好有一间房子（一间房子只能住一对夫妇，住了人的空地别人还是可以走过的）。
目标：使得所有人都结成夫妇且住在房子里（中性人可以和男的结成夫妇，也可以和女的结成夫妇）
所花的最大时间最小，即最后一个人住进房子的时间最小。
思路:
假设没有变性人，则这题就是一个经典建图了，
二分最大时间，然后建图即可。
设now为每个人允许运动的最大时间
男人连源点 flow = 1
男人连能走到的房子 flow = 1 （所谓能走到就是所花时间<=now)
房子连女人 flow = 1
女人连汇点 flow = 1
因为一个房子只能住一个人，所以房子拆点一下，限流为1
而对于变性人其实并不是任意匹配的，当且仅当 abs(男人数量-女人数量)=1时才能找到解，即一开始就能确定变性人的性别。
f2的话一直wa87，==
#include<iostream>#include<stdio.h>#include<string.h>#include<queue>#include<cmath>template <class T>inline bool rd(T &ret) {char c; int sgn;if (c = getchar(), c == EOF) return 0;while (c != '-' && (c<'0' || c>'9')) c = getchar();sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c – '0');while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c – '0');ret *= sgn;return 1;}template <class T>inline void pt(T x) {if (x <0) {putchar('-');x = -x;}if (x>9) pt(x / 10);putchar(x % 10 + '0');}using namespace std;typedef long long ll;const ll inf = 1e10;const int N = 1700;const int M = 10000000;template<class T>struct Max_Flow {int n;int Q[N], sign;int head[N], level[N], cur[N], pre[N];int nxt[M], pnt[M], E;T cap[M];void Init(int n) {this->n = n + 1;E = 0;std::fill(head, head + this->n, -1);}//有向rw 就= 0 void add(int from, int to, T c, T rw) {pnt[E] = to;cap[E] = c;nxt[E] = head[from];head[from] = E++;pnt[E] = from;cap[E] = rw;nxt[E] = head[to];head[to] = E++;}bool Bfs(int s, int t) {sign = t;std::fill(level, level + n, -1);int *front = Q, *tail = Q;*tail++ = t; level[t] = 0;while (front < tail && level[s] == -1) {int u = *front++;for (int e = head[u]; e != -1; e = nxt[e]) {if (cap[e ^ 1] > 0 && level[pnt[e]] < 0) {level[pnt[e]] = level[u] + 1;*tail++ = pnt[e];}}}return level[s] != -1;}void Push(int t, T &flow) {T mi = inf;int p = pre[t];for (int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {mi = std::min(mi, cap[p]);}for (int p = pre[t]; p != -1; p = pre[pnt[p ^ 1]]) {cap[p] -= mi;if (!cap[p]) {sign = pnt[p ^ 1];}cap[p ^ 1] += mi;}flow += mi;}void Dfs(int u, int t, T &flow) {if (u == t) {Push(t, flow);return;}for (int &e = cur[u]; e != -1; e = nxt[e]) {if (cap[e] > 0 && level[u] – 1 == level[pnt[e]]) {pre[pnt[e]] = e;Dfs(pnt[e], t, flow);if (level[sign] > level[u]) {return;}sign = t;}}}T Dinic(int s, int t) {pre[s] = -1;T flow = 0;while (Bfs(s, t)) {std::copy(head, head + n, cur);Dfs(s, t, flow);}return flow;}};Max_Flow <ll>F;struct node{int x, y; ll t;node(int a = 0, int b = 0, ll c = 0) :x(a), y(b), t(c){}}AA[N], B[N], C;int n, m, a, b;char mp[24][24];ll Dis[24][24][24][24];int has1(int x){ return x – 1; }int has2(int x, int y){ return a + (x – 1)*m + y – 1; }int has3(int x, int y){ return a + n*m + (x – 1)*m + y – 1; }int has4(int x){ return a + 2 * n*m + x – 1; }bool ok(ll now){int from = has4(b) + 1, to = from + 1;F.Init(to+10);for (int i = 1; i <= a; i++){F.add(from, has1(i), 1, 0);for (int x = 1; x <= n; x++)for (int y = 1; y <= m; y++)if (Dis[x][y][AA[i].x][AA[i].y] < inf && Dis[x][y][AA[i].x][AA[i].y] * AA[i].t <= now)F.add(has1(i), has2(x, y), 1, 0);}for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)if (mp[i][j] == '.') F.add(has2(i, j), has3(i, j), 1, 0);for (int i = 1; i <= b; i++){F.add(has4(i), to, 1, 0);for (int x = 1; x <= n; x++)for (int y = 1; y <= m; y++)if (Dis[x][y][B[i].x][B[i].y] < inf && Dis[x][y][B[i].x][B[i].y] * B[i].t <= now)F.add(has3(x, y), has4(i), 1, 0);}return F.Dinic(from, to) == b;}ll solve(){if (a != b)return -1;ll ans = -1, l = 0, r = 1e18;//二分答案while (l <= r){ll mid = (l + r) >> 1;if (ok(mid)){r = mid – 1;ans = mid;}elsel = mid + 1;}return ans;}void pre(){for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)for (int x = 1; x <= n; x++)for (int y = 1; y <= m; y++)Dis[i][j][x][y] = inf;int step[4][2] = { 0, 1, 0, -1, 1, 0, -1, 0 };for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++){if (mp[i][j] == '*')continue;Dis[i][j][i][j] = 0;for (int k = 0; k < 4; k++){int x = i + step[k][0], y = j + step[k][1];if (1 <= x && x <= n && 1 <= y&&y <= m && mp[x][y] == '.')Dis[i][j][x][y] = 1;}}//Floydfor (int x1 = 1; x1 <= n; x1++)for (int y1 = 1; y1 <= m; y1++) if (mp[x1][y1] == '.')for (int x2 = 1; x2 <= n; x2++)for (int y2 = 1; y2 <= m; y2++) if (mp[x2][y2]=='.' && Dis[x2][y2][x1][y1]<inf)for (int x3 = 1; x3 <= n; x3++)for (int y3 = 1; y3 <= m; y3++) if (mp[x3][y3] == '.')Dis[x2][y2][x3][y3] = min(Dis[x2][y2][x3][y3], Dis[x2][y2][x1][y1] + Dis[x1][y1][x3][y3]);if (a>b)B[++b] = C; //确定变性人的性别，，就相当于直接把他看成男（女）人else AA[++a] = C;}void input(){rd(n); rd(m); rd(a); rd(b);for (int i = 1; i <= n; i++)scanf("%s", mp[i] + 1);rd(C.x); rd(C.y); rd(C.t);for (int i = 1; i <= a; i++) { rd(AA[i].x); rd(AA[i].y); rd(AA[i].t); }for (int i = 1; i <= b; i++) { rd(B[i].x); rd(B[i].y); rd(B[i].t); }}int main(){input();pre();cout<<solve()<<endl;return 0;}/*1 1 1 0.1 1 11 1 1*/
第一个青春是上帝给的;第二个的青春是*自己努力的
