[BZOJ 1001] 狼抓兔子

描述

http://www.lydsy.com/JudgeOnline/problem.php?id=1001

分析

这是道经典的对偶图问题, 平面图最大流问题可以转化为其对偶图的最短路问题.

转化的方法就是将每个三角形区域看作是一个点, 如果两个三角形区域有公共线, 就在两个结点之间连一条权值为公共线容量的边.

关于编号问题我定义了一个id数组. 表示以点 (x(0~n-2), y(0~m-2)) 为左上角的三角形区域的编号. 右上三角的编号为id[x][y], 右下为id[x][y]^1.

对偶图问题还是不太懂.
一开始我认为应该建立无向图, MLE多次后改为有向图. 但两个三角之间是竖向边的情况该从谁向谁连呢? 两种都试了, 发现竟然都能AC, 但是一个7296ms一个1752ms. 求解释.

代码

139808kb 1752ms // 感觉好弱…

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue> 
#include<algorithm>
using namespace std;

const int INF  = 1e9 + 7;
const int maxn = 2500000 + 10;
const int maxm = 5000 + 10;

struct Edge {
    int from, to, dist;
};

vector<Edge> edges;
vector<int> G[maxn];

void AddEdge(int from, int to, int dist) {
    edges.push_back((Edge){from, to, dist});
    G[from].push_back(edges.size()-1);
}

int s, t, d[maxn];
queue<int> Q;
bool inq[maxn];

int SPFA() {
    for(int i = 0; i <= t; i++) d[i] = INF;
    Q.push(s); inq[s] = 1; d[s] = 0; 

    while(!Q.empty()) {
        int u = Q.front(); Q.pop();
        inq[u] = 0;
        for(int i = 0; i < G[u].size(); i++) {
            Edge& e = edges[G[u][i]];
            if(d[e.to] > d[u] + e.dist) {
                d[e.to] = d[u] + e.dist;
                if(!inq[e.to]) {
                    Q.push(e.to);
                    inq[e.to] = 1;
                }
            }
        }
    }
    return d[t];
}

int id[maxm][maxm]; 
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    // special judge
    if(n == 1 || m == 1) {
        int minv = INF;
        for(int i = 0; i < max(n, m)-1; i++) {
            int v;
            scanf("%d", &v);
            minv = min(minv, v);
        }
        printf("%d\n", minv);
        return 0;
    }
    // 根据左上角的点的编号(x(0~n-2), y(0~m-2))计算格子的编号 
    int c = 0;
    for(int x = 0; x < n-1; x++) {
        for(int y = 0; y < m-1; y++)
            id[x][y] = c, c += 2;
    }
    s = c; t = s^1;
    // 横向道路
    for(int x = 0; x < n; x++)
        for(int y = 0; y < m-1; y++) {
            int v;
            scanf("%d", &v);
            if(x == 0) AddEdge(s, id[x][y], v); 
            else if(x == n-1) AddEdge(id[x-1][y]^1, t, v); 
            else AddEdge(id[x-1][y]^1, id[x][y], v);
        }
    // 纵向道路 
    for(int x = 0; x < n-1; x++)
        for(int y = 0; y < m; y++) {
            int v;
            scanf("%d", &v);
            if(y == 0) AddEdge(id[x][y]^1, t, v);
            else if(y == m-1) AddEdge(s, id[x][y-1], v);
            else AddEdge(id[x][y-1], id[x][y]^1, v); // 连法不同, 效率相差很大！
        }
    // 斜向道路 
    for(int x = 0; x < n-1; x++)
        for(int y = 0; y < m-1; y++) {
            int v;
            scanf("%d", &v);
            AddEdge(id[x][y], id[x][y]^1, v);
        }
    printf("%d\n", SPFA());
    return 0;
}

还有个网络流写法(MLE)
权当娱乐…

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue> 
#include<algorithm>
using namespace std;

const int INF = 1e9 + 7;
const int maxn = 1000000 + 10;

struct Edge {
    int from, to, cap, flow;
};

struct Dinic {
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  bool vis[maxn];
  int d[maxn], cur[maxn]; 

  // 其实如果都是无向边, 反向边的容量可以直接设为cap, 就不用加两次边了
  void AddEdge(int from, int to, int cap) {
    edges.push_back((Edge){from, to, cap, 0});
    edges.push_back((Edge){to, from, 0, 0});
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
  }

  bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    vis[s] = 1;
    d[s] = 0;
    while(!Q.empty()) {
      int x = Q.front(); Q.pop();
      for(int i = 0; i < G[x].size(); i++) {
        Edge& e = edges[G[x][i]];
        if(!vis[e.to] && e.cap > e.flow) {
          vis[e.to] = 1;
          d[e.to] = d[x] + 1;
          Q.push(e.to);
        }
      }
    }
    return vis[t];
  }

  int DFS(int x, int a) {
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int& i = cur[x]; i < G[x].size(); i++) {
      Edge& e = edges[G[x][i]];
      if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap-e.flow))) > 0) {
        e.flow += f;
        edges[G[x][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(a == 0) break;
      }
    }
    return flow;
  }

  int Maxflow(int s, int t) {
    this->s = s; this->t = t;
    int flow = 0;
    while(BFS()) {
      memset(cur, 0, sizeof(cur));
      flow += DFS(s, INF);
    }
    return flow;
  }
}dinic;

int n, m;

inline int encode(int x, int y) {
    return x * m + y + 1;
}

int main() {
    scanf("%d%d", &n, &m);
    int cap;
    for(int x = 0; x < n; x++) 
        for(int y = 0; y < m-1; y++) {
            scanf("%d", &cap);
            dinic.AddEdge(encode(x,y), encode(x, y+1), cap);
            dinic.AddEdge(encode(x, y+1), encode(x,y), cap); // 
        }
    for(int x = 0; x < n-1; x++)
        for(int y = 0; y < m; y++) {
            scanf("%d", &cap);
            dinic.AddEdge(encode(x,y), encode(x+1, y), cap);
            dinic.AddEdge(encode(x+1, y), encode(x,y), cap); // 
        }
    for(int x = 0; x < n-1; x++)
        for(int y = 0; y < m-1; y++) {
            scanf("%d", &cap);
            dinic.AddEdge(encode(x,y), encode(x+1, y+1), cap);
            dinic.AddEdge(encode(x+1, y+1), encode(x,y), cap); // 
        }
    printf("%d\n", dinic.Maxflow(1, m*n));
    return 0;
}