题目链接
简单网络流。
源点向蜥蜴连流量为(1)的边。
能跳出去的点向汇点连流量为(INF)的边。
把每个点拆成(2)个点，(O(n^4))枚举两两点，如果距离小于等于(d)，就互连流量为(INF)的边。
然后跑(dinic)就行了。
#include <cstdio>
#include <queue>
#include <cmath>
#include <cstring>
#define INF 2147483647
using namespace std;
const int N = 25;
const int MAXN = 100010;
const int MAXM = 200010;
inline int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
return s * w;
}
struct Edge{
int next, to, rest;
}e[MAXM];
int s, t, num = 1, n, m, d;
int head[MAXN];
char ch;
inline void Add(int from, int to, int flow){
e[++num] = (Edge){ head[from], to, flow }; head[from] = num;
e[++num] = (Edge){ head[to], from, 0 }; head[to] = num;
}
int level[MAXN], now, tot, h[N][N];
queue <int> q;
int re(){
memset(level, 0, sizeof level);
while(q.size()) q.pop();
q.push(s); level[s] = 1;
while(q.size()){
now = q.front(); q.pop();
for(int i = head[now]; i; i = e[i].next)
if(e[i].rest && !level[e[i].to]){
level[e[i].to] = level[now] + 1;
q.push(e[i].to);
}
}
return level[t];
}
int findflow(int u, int flow){
if(!flow || u == t) return flow;
int f = 0, t;
for(int i = head[u]; i; i = e[i].next){
if(e[i].rest && level[e[i].to] == level[u] + 1){
f += (t = findflow(e[i].to, min(flow - f, e[i].rest)));
e[i].rest -= t; e[i ^ 1].rest += t;
}
}
if(!f) level[u] = 0;
return f;
}
int dinic(){
int ans = 0;
while(re())
ans += findflow(s, INF);
return ans;
}
inline int r(int i, int j){
return (i - 1) * m + j;
}
inline int c(int i, int j){
return r(i, j) + 10000;
}
inline double dis(int i, int j, int k, int l){
return sqrt((k - i) * (k - i) + (l - j) * (l - j));
}
int main(){
n = read(); m = read(); d = read(); s = 2008; t = 2009;
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j){
ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
h[i][j] = ch - '0';
}
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j){
ch = getchar();
while(ch != 'L' && ch != '.') ch = getchar();
if(ch == 'L') Add(s, r(i, j), 1), ++tot;
if(min(min(i, j), min(n - i + 1, m - j + 1)) <= d) Add(c(i, j), t, INF);
Add(r(i, j), c(i, j), h[i][j]);
}
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j)
for(int k = 1; k <= n; ++k)
for(int l = 1; l <= m; ++l)
if(dis(i, j, k, l) <= d)
Add(c(i, j), r(k, l), INF), Add(c(k, l), r(i, j), INF);
printf("%d
", tot - dinic());
return 0;
}