Hdu3046 Pleasant sheep and big big wolf(最小割)

题目链接:点击跳转

可以用网络流解决，难点在于如何建图。
针对每一个点，他们都可以移动到四周的点上，那么对于每个点，向四周的点建一条流量为1的边，对于每一个1的点，建一条源点到这个点的边，流量为inf，对于每一个2的点，建一条这个点到汇点的边，流量为inf。这样子，这个问题就被转化为最大流问题，因为最大流=最小割，那么最后得到的就是题目要求的答案。
代码如下:

#include<bits/stdc++.h>

using namespace std;
typedef unsigned long long ll;
const int MAXN = 5e5 + 5e2;
const int INF = 0x3f3f3f3f;
#define endl '\n'

inline void IO_STREAM() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
}

struct Edge{
    int to,val,nxt;
}e[MAXN];

int head[MAXN], deep[MAXN], gra[205][205], to[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
int n, m, s, t, cnt;
ll res;

inline void init() {
    memset(head, -1, sizeof(head));
    cnt = 0;
    res = 0;
}

inline void add(int from, int to, int val) {
    e[cnt] = Edge{to, val, head[from]};
    head[from] = cnt++;
}

inline bool check(int x, int y) {
    if (x >= 0 && x < n && y >= 0 && y < m) return true;
    return false;
}

inline void addPath(int x, int y) {
    for (int i = 0; i < 4; i++) {
        int xx = x + to[i][0];
        int yy = y + to[i][1];
        if (check(xx, yy)) {
            add(xx * m + yy + 1, x * m + y + 1, 1);
            add(x * m + y + 1, xx * m + yy + 1, 0);
        }
    }
    if (gra[x][y] == 1) {
        add(x * m + y + 1, t, INF);
        add(t, x * m + y + 1, 0);
    } else if (gra[x][y] == 2){
        add(s, x * m + y + 1, INF);
        add(x * m + y + 1, s, 0);
    }
}

inline bool bfs() {
    memset(deep, -1, sizeof(deep));
    deep[s] = 1;
    queue<int>q;
    q.push(s);
    while (!q.empty()) {
        int front = q.front();
        q.pop();
        for (int i = head[front]; i != -1; i = e[i].nxt) {
            if (deep[e[i].to] == -1 && e[i].val) {
                deep[e[i].to] = deep[front] + 1;
                q.push(e[i].to);
            }
        }
    }
    return (deep[t] != -1);
}

inline int dfs(int x, int totFlow) {
    if (x == t || totFlow ==0) return totFlow;
    int ans = totFlow;
    for (int i = head[x]; i != -1; i = e[i].nxt) {
        if (deep[e[i].to] == deep[x] + 1 && e[i].val) {
            int flow = dfs(e[i].to, min(totFlow, e[i].val));
            e[i].val -= flow;
            e[i ^ 1].val += flow;
            totFlow -= flow;
            if (totFlow == 0) return ans;
        }
    }
    return ans - totFlow;
}

inline void dinic() {
    while (bfs()) {
        res += dfs(s, INF);
    }
}

int main() {
    IO_STREAM();
    int Case = 1;
    while (cin >> n >> m) {
        init();
        s = 0, t = n * m + 5;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                cin >> gra[i][j];
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                addPath(i, j);
            }
        }
        dinic();
        cout << "Case " << Case++ << ":" << endl;
        cout << res << endl;
//        cout.flush(); //debug
    }
    return 0;
}