UVALive - 3661 Animal Run （平面图+最小割+对偶图+最短路）

题目大意：有很多只小动物要从左上角跑到右下角，给出每条线路所需的人手，问至少需要多少人手，才能将所有动物抓住

解题思路：最小割，就是最小割，但是用最大流处理不了，边太多了
具体可以参考算法合集之《浅析最大最小定理在信息学竞赛中的应用》
知道了这个后，这题估计就可以解了
给出我的建图方式
将每一个小三角形从左往右，从上到下依次编号为1-2-3。。
每行的同一个三角行的编号差就是2 * (m - 1)

如图

#include 
#include 
#include 
#include 
using namespace std;
const int N = 1111 * 1111 * 3;
#define INF 0x3f3f3f3f

struct Edge{
    int to, next, dist;
}E[N<<2];

struct Node {
    int id, val;
    Node(int id, int val): id(id), val(val) {}
    bool operator < (const Node &a) const {
        return val > a.val;
    }
};

int head[N], d[N];
int tot, source, sink, n, m;
bool inq[N];

void AddEdge(int from, int to, int dist) {
    E[tot].to = to;
    E[tot].dist = dist;
    E[tot].next = head[from];
    head[from] = tot++;
}

void init() {
    memset(head, -1, sizeof(head));
    tot = 0;
    source = 2 * (m - 1) * (n - 1) + 2;
    sink = source - 1;

    int u, v, d;
    for (int i = 1; i <= n; i++) 
        for (int j = 1; j <= m - 1; j++) {
            scanf("%d", &d);
            if (i == 1) {
                u = j * 2;
                AddEdge(sink, u, d);
                AddEdge(u, sink, d);
            }
            else if(i == n) {
                u = 2 * (m - 1) * (i - 2) + 2 * (j - 1) + 1;
                AddEdge(source, u, d);
                AddEdge(u, source, d);
            }
            else {
                u = 2 * (m - 1) * (i - 2) + 2 * (j - 1) + 1;
                v = u + 2 * (m - 1) + 1;
                AddEdge(u, v, d);
                AddEdge(v, u, d);
            }
        }

    for (int i = 1; i <= n - 1; i++)
        for (int j = 1; j <= m; j++) {
            scanf("%d", &d);
            if (j == 1) {
                u = 2 * (m - 1) * (i - 1) + 1;
                AddEdge(u, source, d);
                AddEdge(source, u, d);
            }
            else if(j == m) {
                u = 2 * (m - 1) * i;
                AddEdge(u, sink, d);
                AddEdge(sink, u, d);
            }
            else {
                u = 2 * (m - 1) * (i - 1) + 2 * (j - 1); 
                v = u + 1;
                AddEdge(u, v, d);
                AddEdge(v, u, d);
            }
        }

    for (int i = 1; i <= n - 1; i++)
        for (int j = 1; j <= m - 1; j++) {
            scanf("%d", &d);
            u = 2 * (m - 1) * (i - 1) + 2 * (j - 1) + 1;
            v = u + 1;
            AddEdge(u, v, d);
            AddEdge(v, u, d);
        }
}

void BFS() {
    priority_queue q;
    memset(inq, 0, sizeof(inq)); 
    for (int i = 1; i <= sink; i++)
        d[i] = INF;
    int u;
    d[source] = 0;
    q.push(Node(source, 0));
    while (!q.empty()) {
        Node cur = q.top();
        q.pop();

        int u = cur.id;
        if (inq[u])
            continue;
        inq[u] = true;;
        for (int i = head[u]; i != -1; i = E[i].next) {
            int v = E[i].to;
            if (d[v] > d[u] + E[i].dist) {
                d[v] = d[u] + E[i].dist;
                q.push(Node(v, d[v]));
            }
        }
    }
}

int main() {
    int cas = 1;
    while (scanf("%d%d", &n, &m) != EOF && n + m) {
        init();
        BFS();
        printf("Case %d: Minimum = %d\n", cas++, d[sink]);
    }
    return 0;
}