方格取数问题 最小割

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

11

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;
}

原文地址：https://www.cnblogs.com/zxyqzy/p/10353555.html