Project Euler 345: Matrix Sum

题目

思路：

将问题转化成最小费用流

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "
";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 200 + 10;
const int INF = 0x3f3f3f3f;
int a[N][N];
struct edge {
    int to, cap, cost, rev;
};
int V;
vector<edge>g[N];
int h[N], dis[N], prevv[N], preve[N];
void add_edge(int u, int v, int cap, int cost) {
    g[u].pb({v, cap, cost, g[v].size()});
    g[v].pb({u, 0, -cost, g[u].size()-1});
}
int min_cost_flow(int s, int t, int f) {
    int res = 0;
    mem(h, 0);
    while(f > 0) {
        priority_queue<pii, vector<pii>, greater<pii> > q;
        mem(dis, 0x3f);
        dis[s] = 0;
        q.push({0, s});
        while(!q.empty()) {
            pii p = q.top();
            q.pop();
            int v = p.se;
            if(dis[v] < p.fi) continue;
            for (int i = 0; i < g[v].size(); ++i) {
                edge &e = g[v][i];
                if(e.cap > 0 && dis[e.to] > dis[v] + e.cost + h[v] - h[e.to]) {
                    dis[e.to] = dis[v] + e.cost + h[v] - h[e.to];
                    prevv[e.to] = v;
                    preve[e.to] = i;
                    q.push({dis[e.to], e.to});
                }
            }
        }
        if(dis[t] == INF) return -1;
        for (int v = 0; v < V; ++v) h[v] += dis[v];
        int d = f;
        for (int v = t; v != s; v = prevv[v]) d = min(d, g[prevv[v]][preve[v]].cap);
        f -= d;
        res += d*h[t];
        for (int v = t; v != s; v = prevv[v]) {
            edge &e = g[prevv[v]][preve[v]];
            e.cap -= d;
            g[v][e.rev].cap += d;
        }
    }
    return res;
}
int main() {
    int n = 15;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            scanf("%d", &a[i][j]);
            add_edge(i, j+n, 1, 1000-a[i][j]);
        }
    }
    int s = 0, t = n+n+1;
    V = t+1;
    for (int i = 1; i <= n; ++i) add_edge(s, i, 1, 0);
    for (int i = 1; i <= n; ++i) add_edge(i+n, t, 1, 0);
    cout << 1000*n - min_cost_flow(s, t, n);
    return 0;
}