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; }
时间: 2024-10-06 00:53:38