BZOJ1001 [BeiJing2006]狼抓兔子

一眼最小割，转化成最大流来做。

然后发现点数达到10^6级别，妥妥TLE，于是需要进一步思考。

由网上大量题解可知，一个图的最大流等于它的对偶图的最短路，于是只要Dijkstra就可以了。

建图有点恶心。。。查了好长时间。。。

/**************************************************************
    Problem: 1001
    User: rausen
    Language: C++
    Result: Accepted
    Time:520 ms
    Memory:94588 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
 
using namespace std;
 
struct edges{
    int next, to, v;
} e[6000005];
 
struct heap_node{
    int v, to;
};
inline bool operator < (const heap_node &a, const heap_node &b){
    return a.v > b.v;
}
 
priority_queue <heap_node> h;
heap_node NODE;
int tot, S, T, n, m;
int d[3000005], first[3000005];
char ch;
 
inline int read(){
    int x = 0, sgn = 1;
    ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-') sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return sgn * x;
}
 
inline void add_edge(int x, int y, int v){
    e[++tot].next = first[x], first[x] = tot;
    e[tot].to = y, e[tot].v = v;
}
 
void add_Edges(const int x, const int y, const int v){
    add_edge(x, y, v);
    add_edge(y, x, v);
}
 
inline void add_to_heap(const int p){
    for (int x = first[p]; x; x = e[x].next)
        if (d[e[x].to] == -1){
            NODE.v = e[x].v + d[p], NODE.to = e[x].to;
            h.push(NODE);
        }
}
 
int Dijkstra(int S, int T){
    memset(d, -1, sizeof(d));
    while (!h.empty()) h.pop();
    d[S] = 0, add_to_heap(S);
    int p;
    while (d[T] == -1){
        while (d[h.top().to] != -1) h.pop();
        p = h.top().to;
        d[p] = h.top().v;
        h.pop();
        add_to_heap(p);
    }
    return d[T];
}
 
void build_graph(){
    int Cost, x, y;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j < m; ++j){
            Cost = read();
            x = i == 1 ? S : (2 * i - 3) * (m - 1) + j;
            y = i == n ? T : (2 * i - 2) * (m - 1) + j;
            add_Edges(x, y, Cost);
        }
    for (int i = 1; i < n; ++i)
        for (int j = 1; j <= m; ++j){
            Cost = read();
            x = j == 1 ? T : 2 * (i - 1) * (m - 1) + j - 1;
            y = j == m ? S : 2 * (i - 1) * (m - 1) + j - 1 + m;
            add_Edges(x, y, Cost);
        }
    for (int i = 1; i < n; ++i)
        for (int j = 1; j < m; ++j){
            Cost = read();
            x = (2 * i - 2) * (m - 1) + j;
            y = (2 * i - 1) * (m - 1) + j;
            add_Edges(x, y, Cost);
        }
}
 
int main(){
    n = read(), m = read();
    if (n == 1 && m == 1){
        printf("%d
", 0);
        return 0;
    }
     
    S = 0, T = 2 * (m - 1) * (n - 1) + 1;
    build_graph();         
    printf("%d
", Dijkstra(S, T));
    return 0;
}