方格取数问题最小割

末蓝、 2021-12-16 20:04 311阅读 0赞

## 题目背景 ##

none!

## 题目描述 ##

在一个有 m\*n 个方格的棋盘中，每个方格中有一个正整数。现要从方格中取数，使任意 2 个数所在方格没有公共边，且取出的数的总和最大。试设计一个满足要求的取数算法。对于给定的方格棋盘，按照取数要求编程找出总和最大的数。

## 输入输出格式 ##

**输入格式：**

第 1 行有 2 个正整数 m 和 n，分别表示棋盘的行数和列数。接下来的 m 行，每行有 n 个正整数，表示棋盘方格中的数。

**输出格式：**

程序运行结束时，将取数的最大总和输出

## 输入输出样例 ##

**输入样例\#1：** 复制

3 3
    1 2 3
    3 2 3
    2 3 1

**输出样例\#1：** 复制

## 说明 ##

m,n<=100

**总和sum- dinic()；**

#include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<map>
    #include<set>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<ctime>
    #include<deque>
    #include<stack>
    #include<functional>
    #include<sstream>
    //#include<cctype>
    //#pragma GCC optimize(2)
    using namespace std;
    #define maxn 100005
    #define inf 0x7fffffff
    //#define INF 1e18
    #define rdint(x) scanf("%d",&x)
    #define rdllt(x) scanf("%lld",&x)
    #define rdult(x) scanf("%lu",&x)
    #define rdlf(x) scanf("%lf",&x)
    #define rdstr(x) scanf("%s",x)
    typedef long long  ll;
    typedef unsigned long long ull;
    typedef unsigned int U;
    #define ms(x) memset((x),0,sizeof(x))
    const long long int mod = 1e9;
    #define Mod 1000000000
    #define sq(x) (x)*(x)
    #define eps 1e-5
    typedef pair<int, int> pii;
    #define pi acos(-1.0)
    //const int N = 1005;
    #define REP(i,n) for(int i=0;i<(n);i++)
    typedef pair<int, int> pii;
    
    inline int rd() {
    	int x = 0;
    	char c = getchar();
    	bool f = false;
    	while (!isdigit(c)) {
    		if (c == '-') f = true;
    		c = getchar();
    	}
    	while (isdigit(c)) {
    		x = (x << 1) + (x << 3) + (c ^ 48);
    		c = getchar();
    	}
    	return f ? -x : x;
    }
    
    
    ll gcd(ll a, ll b) {
    	return b == 0 ? a : gcd(b, a%b);
    }
    int sqr(int x) { return x * x; }
    
    
    
    /*ll ans;
    ll exgcd(ll a, ll b, ll &x, ll &y) {
    	if (!b) {
    		x = 1; y = 0; return a;
    	}
    	ans = exgcd(b, a%b, x, y);
    	ll t = x; x = y; y = t - a / b * y;
    	return ans;
    }
    */
    
    int n, m;
    int st, ed;
    struct node {
    	int u, v, nxt, w;
    }edge[maxn << 1];
    
    int head[maxn], cnt;
    
    void addedge(int u, int v, int w) {
    	edge[cnt].u = u; edge[cnt].v = v; edge[cnt].nxt = head[u];
    	edge[cnt].w = w; head[u] = cnt++;
    }
    
    int rk[maxn];
    
    int bfs() {
    	queue<int>q;
    	ms(rk);
    	rk[st] = 1;
    	q.push(st);
    	while (!q.empty()) {
    		int tmp = q.front(); q.pop();
    		for (int i = head[tmp]; i != -1; i = edge[i].nxt) {
    			int to = edge[i].v;
    			if (rk[to] || edge[i].w <= 0)continue;
    			rk[to] = rk[tmp] + 1; q.push(to);
    		}
    	}
    	return rk[ed];
    }
    
    int dfs(int u, int flow) {
    	if (u == ed)return flow;
    	int add = 0;
    	for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) {
    		int v = edge[i].v;
    		if (rk[v] != rk[u] + 1 || !edge[i].w)continue;
    		int tmpadd = dfs(v, min(edge[i].w, flow - add));
    		if (!tmpadd) { rk[v] = -1; continue; }
    		edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd;
    		add += tmpadd;
    	}
    	return add;
    }
    
    int ans;
    void dinic() {
    	while (bfs())ans += dfs(st, inf);
    }
    //int n, m;
    int a[200][200];
    int dx[] = { 0,0,-1,1 };
    int dy[] = { 1,-1,0,0 };
    bool OK(int x, int y) {
    	return x >= 1 && x <= n && y >= 1 && y <= m;
    }
    int getpos(int x, int y) {
    	return (x - 1)*m + y;
    }
    int main()
    {
    	//	ios::sync_with_stdio(0);
    	n = rd(); m = rd(); memset(head, -1, sizeof(head));
    	int sum = 0;
    	for (int i = 1; i <= n; i++) {
    		for (int j = 1; j <= m; j++)a[i][j] = rd(), sum += a[i][j];
    	}
    	st = 0; ed = n * m + 4;
    	for (int i = 1; i <= n; i++) {
    		for (int j = 1; j <= m; j++) {
    			if ((i + j) % 2)addedge(st, getpos(i, j), a[i][j]), addedge(getpos(i, j), st, a[i][j]);
    			else addedge(getpos(i, j), ed, a[i][j]), addedge(ed, getpos(i, j), 0);
    		}
    	}
    	for (int i = 1; i <= n; i++) {
    		for (int j = 1; j <= m; j++) {
    			if ((i + j) % 2) {
    				for (int k = 0; k < 4; k++) {
    					int nx = i + dx[k];
    					int ny = j + dy[k];
    					if (OK(nx, ny))addedge(getpos(i, j), getpos(nx, ny), inf), addedge(getpos(nx, ny), getpos(i, j), 0);
    				}
    			}
    		}
    	}
    	//cout << 1 << endl;
    	dinic();
    	
    	printf("%d\n", sum - ans);
    	return 0;
    }

转载于:https://www.cnblogs.com/zxyqzy/p/10353555.html