【網路流 24 題】方格取數(二分圖的最大點權獨立集)
阿新 • • 發佈:2019-02-17
題意
在一個有 個方格的棋盤中,每個方格中有一個正整數。
現要從方格中取數,使任意 個數所在方格沒有公共邊,且取出的數的總和最大。試設計一個滿足要求的取數演算法。
題解
題目要求不相鄰,可以轉換為最大獨立集,又由於點權不全為1,則為最大點權獨立集。
最大點權獨立集 = 總點權 - 最小點權覆蓋,問題轉換為如何求最小點權覆蓋。
建立二分圖,源點向左點集連邊, 容量為點權。右點集向匯點連邊,容量為點權。內部的點連邊,容量為INF。跑最大流,求出最小割即可。
程式碼
#include<bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 1e6+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
int n, m;
int mp[50][50];
struct Dinic {
int head[nmax], cur[nmax], d[nmax];
bool vis[nmax] ;
int tot, n, m, s, t, front, tail;
int qqq[nmax];
struct edge {
int nxt, to, w, cap, flow;
} e[nmax<<1];
void init(int n) {
this->n = n;
memset(head, -1, sizeof head);
memset(cur, 0, sizeof cur);
memset(e,0,sizeof e);
this-> tot = 0;
}
int add_edge(int u, int v, int c) {
int temp = tot;
e[tot].to = v, e[tot].cap = c, e[tot].flow = 0;
e[tot].nxt = head[u];
head[u] = tot++;
e[tot].to = u, e[tot].cap = c, e[tot].flow = c;
e[tot].nxt = head[v];
head[v] = tot++;
// printf("add %d %d %d\n", u, v, c);
return temp;
}
bool BFS() {
for(int i = 0; i <= n; ++i) vis[i] = false;
front = tail = 0;
vis[s] = 1; d[s] = 0;
qqq[tail++] = s;
while (front < tail) {
int u = qqq[front++];
for (int i = head[u]; i != -1; i = e[i].nxt) {
int v = e[i].to;
if (!vis[v] && e[i].cap > e[i].flow) {
vis[v] = 1;
d[v] = d[u] + 1;
qqq[tail++] = v;
}
}
}
return vis[t];
}
int DFS(int x, int a) {
if (x == t || a == 0) return a;
int Flow = 0, f;
for (int& i = cur[x]; i != -1; i = e[i].nxt) {
int v = e[i].to;
if (d[v] == d[x] + 1 && (f = DFS(v, min(a, e[i].cap - e[i].flow))) > 0) {
Flow += f;
e[i].flow += f;
e[i ^ 1].flow -= f;
a -= f;
if (a == 0) break;
}
}
return Flow;
}
int Maxflow(int s, int t) {
this->s = s, this->t = t;
int Flow = 0;
while (BFS()) {
for (int i = 0; i <= n; i++) cur[i] = head[i];
Flow += DFS(s,INF);
}
return Flow;
}
} dinic;
int main(){
scanf("%d %d", &n, &m);
int sum = 0;
int s = 0, t = n * m + 1;
dinic.init(t);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
scanf("%d", &mp[i][j]);
sum += mp[i][j];
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if ((i + j) % 2 == 1) {
dinic.add_edge(s, (i - 1) * m + j, mp[i][j]);
if (i != n) {
dinic.add_edge((i - 1) * m + j, i * m + j, INF);
}
if (j != m) {
dinic.add_edge((i - 1) * m + j, (i - 1) * m + j + 1, INF);
}
if (i != 1) {
dinic.add_edge((i - 1) * m + j, (i - 2) * m + j, INF);
}
if (j != 1) {
dinic.add_edge((i - 1) * m + j, (i - 1) * m + j - 1, INF);
}
} else {
dinic.add_edge((i - 1) * m + j, t, mp[i][j]);
}
}
}
int mxflow = dinic.Maxflow(s, t);
// printf("%d\n", mxflow);
printf("%d\n", sum - mxflow);
return 0;
}