Leapin' Lizards 【HDU - 2732】【最大流】

题目链接

---

  题意：就是给你这样的N*M的点（其中M还没有给出，但是M可以算得），然后先输出N、D，N行，D是每个青蛙能跳的距离（欧拉距离，以为是曼哈顿距离的我WA了两发…… QAQ），然后是第一张N*M的图，代表着的是每个点能被起跳的次数，从它起跳需要消耗一次，然后第二幅N*M的图代表的是是否有青蛙。

  直接最大流就是了，但是这里的D可能会>3，其他人错的原因，我完全是被卡了欧拉距离。

---

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define esp 1e-6
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int S = 0, maxN = 1e3 + 7, maxE = 4e4 + 7;
int N, M, D, head[maxN], cur[maxN], cnt, T, all;
#define _id(x, y) (x - 1) * M + y
char num[25][25], lea[25][25];
struct Eddge
{
    int nex, to, flow;
    Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), flow(c) {}
}edge[maxE];
inline void addEddge(int u, int v, int flow)
{
    edge[cnt] = Eddge(head[u], v, flow);
    head[u] = cnt++;
}
inline void _add(int u, int v, int flow) { addEddge(u, v, flow); addEddge(v, u, 0); }
bool check(int x, int y) { return x <= D || y <= D || N - x < D || M - y < D; }
bool In_Map(int x, int y) { return x >= 1 && y >= 1 && x <= N && y <= M; }
bool _link(int x1, int y1, int x2, int y2) { return sqrt( (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) ) <= D; }
int deep[maxN];
queue<int> Q;
bool bfs()
{
    memset(deep, 0, sizeof(deep));  while(!Q.empty()) Q.pop();
    Q.push(S);  deep[S] = 1;
    while(!Q.empty())
    {
        int u = Q.front();  Q.pop();
        for(int i=head[u], v, f; ~i; i=edge[i].nex)
        {
            v = edge[i].to; f = edge[i].flow;
            if(f && !deep[v])
            {
                deep[v] = deep[u] + 1;
                Q.push(v);
            }
        }
    }
    return deep[T];
}
int dfs(int u, int dist)
{
    if(u == T) return dist;
    for(int &i=cur[u], v, f, di; ~i; i=edge[i].nex)
    {
        v = edge[i].to; f = edge[i].flow;
        if(f && deep[v] == deep[u] + 1)
        {
            di = dfs(v, min(dist, f));
            if(di)
            {
                edge[i].flow -= di;
                edge[i^1].flow += di;
                return di;
            }
        }
    }
    return 0;
}
int Dinic()
{
    int ans = 0, tmp;
    while(bfs())
    {
        memcpy(cur, head, sizeof(cur));
        while((tmp = dfs(S, INF))) ans += tmp;
    }
    return ans;
}
inline void init()
{
    cnt = all = 0;    T = N * M + 401;
    memset(head, -1, sizeof(head));
}
int main()
{
    int tt;  scanf("%d", &tt);
    for(int Cas=1; Cas<=tt; Cas++)
    {
        scanf("%d%d", &N, &D);
        for(int i=1; i<=N; i++) scanf("%s", num[i] + 1);
        for(int i=1; i<=N; i++) scanf("%s", lea[i] + 1);
        M = (int)strlen(num[1] + 1);
        init();
        for(int i=1; i<=N; i++)
        {
            for(int j=1; j<=M; j++)
            {
                if(num[i][j] > '0') _add(_id(i, j), _id(i, j) + 400, num[i][j] - '0');
                if(check(i, j)) _add(_id(i, j) + 400, T, INF);
                /*for(int dx = 0; dx <= D; dx++)
                {
                    for(int dy = 0; dy <= D - dx; dy++)
                    {
                        if(!dx && !dy) continue;
                        if(In_Map(i + dx, j + dy)) _add(_id(i, j) + 400, _id(i + dx, j + dy), INF);
                        if(In_Map(i - dx, j - dy)) _add(_id(i, j) + 400, _id(i - dx, j - dy), INF);
                        if(In_Map(i - dx, j + dy)) _add(_id(i, j) + 400, _id(i - dx, j + dy), INF);
                        if(In_Map(i + dx, j - dy)) _add(_id(i, j) + 400, _id(i + dx, j - dy), INF);
                    }
                }*/
                for(int dx=1; dx<=N; dx++) for(int dy=1; dy<=M; dy++)
                {
                    if(dx == i && dy == j) continue;
                    if(_link(i, j, dx, dy)) _add(_id(i, j) + 400, _id(dx, dy), INF);
                }
            }
        }
        for(int i=1; i<=N; i++)
        {
            for(int j=1; j<=M; j++)
            {
                if(lea[i][j] == 'L')
                {
                    _add(S, _id(i, j), 1);
                    all++;
                }
            }
        }
        int ans = all - Dinic();
        printf("Case #%d: ", Cas);
        if(!ans) printf("no lizard was left behind.\n");
        else if(ans == 1) printf("1 lizard was left behind.\n");
        else printf("%d lizards were left behind.\n", ans);
    }
    return 0;
}