[SCOI2007]蜥蜴

嘟嘟嘟

 

这题主要考网络流建图。

首先每一只蜥蜴像自己所在的石柱连边，然后如果这一只蜥蜴像能跳到的所有石柱连边，那就错了。因为每一只蜥蜴能跳到别的石柱上，然后再跳到更远的石柱上，而上述连边的意思是蜥蜴只能跳到他一步能达到的石柱上。所以，正确的连边应该是以每一个石柱，向能到达的石柱连边，容量为INF。

题中还有一点：每离开一个石柱，石柱高度减一，这相当于限制了每一个石柱能承受的蜥蜴数量。那么一个常见的做法就是拆点：将每一个石柱拆成跳到他，和从他跳走的，然后连一条容量为石柱高度的边。

最后就是对于每一个能从他跳出边界的石柱，从他跳走的点向汇点连一条INF的边。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1205;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

int n, m, t, d, tot = 0;

struct Edge
{
    int from, to, cap, flow;
};
vector<Edge> edges;
vector<int> G[maxn];
void addEdge(int from, int to, int w)
{
    edges.push_back((Edge){from, to, w, 0});
    edges.push_back((Edge){to, from, 0, 0});
    int sz = edges.size(); 
    G[from].push_back(sz - 2);
    G[to].push_back(sz - 1);
}

int dx[20], dy[20], st = 0;
void init(int x)
{
    for(int i = -d; i <= d; ++i)
        for(int j = -d; j <= d; ++j)
            if(i * i + j * j <= d * d) dx[++st] = i, dy[st] = j;
}

int getnum(int x, int y)
{
    return m * (x - 1) + y;
}

bool judge(int x, int y)
{
    if(x + d > n || x - d <= 0 || y + d > m || y - d <= 0) return 1;
    return 0;
}

void add(int x, int y)
{
    int u = getnum(x, y) + n * m;
    for(int i = 1; i <= st; ++i)
    {
        int newx = x + dx[i], newy = y + dy[i];
        if(newx > 0 && newx <= n && newy > 0 && newy <= m) 
            addEdge(u, getnum(newx, newy), INF);
    }
    
}

int dis[maxn];
bool bfs()
{
    Mem(dis, 0); dis[0] = 1;
    queue<int> q; q.push(0);
    while(!q.empty())
    {
        int now = q.front(); q.pop();
        for(int i = 0; i < (int)G[now].size(); ++i)
        {
            Edge& e = edges[G[now][i]];
            if(!dis[e.to] && e.cap > e.flow)
            {
                dis[e.to] = dis[now] + 1;
                q.push(e.to);
            }
        }
    }
    return dis[t];
}
int cur[maxn];
int dfs(int now, int res)
{
    if(now == t || res == 0) return res;
    int flow = 0, f;
    for(int& i = cur[now]; i < (int)G[now].size(); ++i)
    {
        Edge& e = edges[G[now][i]];
        if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(res, e.cap - e.flow))) > 0)
        {
            e.flow += f;
            edges[G[now][i] ^ 1].flow -= f;
            flow += f;
            res -= f;
            if(res == 0) break;
        }
    }
    return flow;
}

int maxflow()
{
    int flow = 0;
    while(bfs())
    {
        Mem(cur, 0);
        flow += dfs(0, INF);
    }
    return flow;
}

char c[25];

int main()
{
    n = read(); m = read(); d = read();
    init(d);
    t = n * m * 3 + 1;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j)
        {
            int x; scanf("%1d", &x);
            if(x)
            {
                int u = getnum(i, j);
                addEdge(u, n * m + u, x);
                if(judge(i, j)) addEdge(n * m + u, t, INF);
                add(i, j);
            }
        }
    for(int i = 1; i <= n; ++i)
    {
        scanf("%s", c + 1);
        for(int j = 1; j <= m; ++j)
            if(c[j] == 'L')
            {
                tot++;
                int u = getnum(i, j) + ((n * m) << 1);
                addEdge(0, u, 1); 
                addEdge(u, getnum(i, j), 1);
            }
    }
    write(tot - maxflow()); enter;    
    return 0;
}