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;
- }
来源: http://www.bubuko.com/infodetail-2945875.html