BZOJ 1066 SCOI 2007 蜥蜴 最大流

题目大意：给出一张图，每一个点有一个寿命，当有这个寿命值个蜥蜴经过后这个点就会消失，一个蜥蜴可以跳到距离不超过d的点上，问最少有多少只蜥蜴无法跳出这张图。

思路：我们将每个点拆点，然后限制流量为这个点的寿命，之后源点向每个蜥蜴连边，互相能够到达的点之间连边，能够跳出这个图的点和汇点连边，跑最大流就是这个图中最多能够跑出去的蜥蜴数量，最后在用总数减去就是最少不能逃出去的数量。

CODE：

#include <cmath>#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 50#define MAXP 1010#define MAXE 5000010#define INF 0x3f3f3f3f#define S 0#define T ((m * n << 1) + 2)using namespace std;int m,n,d,Ts;int src[MAX][MAX],num[MAX][MAX],cnt;char s[MAX];int head[MAXP],total = 1;int next[MAXE],aim[MAXE],flow[MAXE];int deep[MAXP];inline void Add(int x,int y,int f){next[++total] = head[x];aim[total] = y;flow[total] = f;head[x] = total;}inline void Insert(int x,int y,int f){Add(x,y,f);Add(y,x,0);}inline double Calc(int x1,int y1,int x2,int y2){return sqrt((double)(x1 - x2) * (x1 - x2) + (double)(y1 - y2) * (y1 - y2));}inline bool BFS(){static queue<int> q;while(!q.empty())q.pop();memset(deep,0,sizeof(deep));deep[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]]) {deep[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(deep[aim[i]] == deep[x] + 1 && flow[i] && temp) {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;}int main(){cin >> m >> n >> d;for(int i = 1; i <= m; ++i)for(int j = 1; j <= n; ++j) {scanf("%1d",&src[i][j]),num[i][j] = ++cnt;if(src[i][j])Insert(num[i][j] << 1,num[i][j] << 1|1,src[i][j]);}for(int i = 1; i <= m; ++i) {scanf("%s",s + 1);for(int j = 1; j <= n; ++j)if(s[j] == 'L')Insert(S,num[i][j] << 1,1),++Ts;}for(int i = 1; i <= m; ++i)for(int j = 1; j <= n; ++j)for(int _i = 1; _i <= m; ++_i)for(int _j = 1; _j <= n; ++_j) {if(i == _i && j == _j)continue;if(Calc(i,j,_i,_j) > d)continue;Insert(num[i][j] << 1|1,num[_i][_j] << 1,INF);}for(int i = 1; i <= m; ++i)for(int j = 1; j <= n; ++j)if(i <= d || j <= d || m - i < d || n - j < d)Insert(num[i][j] << 1|1,T,INF);int max_flow = 0;while(BFS())max_flow += Dinic(S,INF);cout << Ts - max_flow << endl;return 0;}