[SWUST1753] 分配问题（费用流，最优匹配）

题目链接：https://www.oj.swust.edu.cn/problem/show/1753

由于每一个人只能做一件工作，所以要在源汇点处设置容量为1费用为0，在二分图中间设置容量为inf。

而不是源点到人处设置容量为inf。

其实就是最优匹配问题，费用流，这样建图之后权值正负各跑一遍就行了。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef struct Node {
    int u, v, next;
    LL c, w;
}Node;
const int maxn = 440;
const int maxm = 80010;
const LL mod = 0x3f3f3f3fLL;
const LL inf = (1LL<<55);
int tot, head[maxn];
LL dist[maxn];
LL cost, flow;
Node e[maxm];
int pre[maxn];
bool visit[maxn];
queue<int> Q;
int S, T, N;

void init() {
    S = T = N = 0;
    memset(head, -1, sizeof(head));
    tot = 0;
}

void adde(int u, int v, LL c, LL w) {
    e[tot].u = u; e[tot].v = v; e[tot].c = c; e[tot].w = w; e[tot].next = head[u]; head[u] = tot++;
    e[tot].u = v; e[tot].v = u; e[tot].c = 0; e[tot].w = -w; e[tot].next = head[v]; head[v] = tot++;
}
bool spfa(int s, int t, int n) {
    int i;
    for(i = 0; i <= n; i++) {
        dist[i] = inf;
        visit[i] = 0;
        pre[i] = -1;
    }
    while(!Q.empty()) Q.pop();
    Q.push(s);
    visit[s] = true;
    dist[s] = 0;
    pre[s] = -1;
    while(!Q.empty()) {
        int u = Q.front();
        visit[u] = false;
        Q.pop();
        for(int j = head[u]; j != -1; j = e[j].next) {
            if(e[j].c > 0 && dist[u] + e[j].w < dist[e[j].v]) {
                dist[e[j].v] = dist[u] + e[j].w;
                pre[e[j].v] = j;
                if(!visit[e[j].v]) {
                    Q.push(e[j].v);
                    visit[e[j].v] = true;
                }
            }
        }
    }
    if(dist[t] == inf) return false;
    else return true;
}
LL ChangeFlow(int t) {
    LL det = mod;
    int u = t;
    while(~pre[u]) {
        u = pre[u];
        det = min(det, e[u].c);
        u = e[u].u;
    }
    u = t;
    while(~pre[u]) {
        u = pre[u];
        e[u].c -= det;
        e[u ^ 1].c += det;
        u = e[u].u;
    }
    return det;
}
LL MinCostFlow(int s, int t, int n) {
    LL mincost, maxflow;
    mincost = maxflow = 0;
    while(spfa(s, t, n)) {
        LL det = ChangeFlow(t);
        mincost += det * dist[t];
        maxflow += det;
    }
    cost = mincost;
    flow = maxflow;
    return mincost;
}

int n;
int w[maxn][maxn];

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d", &n)) {
        init();
        S = 0, T = n * 2 + 1, N = T + 1;
        for(int i = 1; i <= n; i++) {
            adde(S, i, 1, 0);
            adde(i+n, T, 1, 0);
            for(int j = 1; j <= n; j++) {
                scanf("%d", &w[i][j]);
                adde(i, j+n, inf, w[i][j]);
            }
        }
        cout << MinCostFlow(S, T, N) << endl;
        init();
        S = 0, T = n * 2 + 1, N = T + 1;
        for(int i = 1; i <= n; i++) {
            adde(S, i, 1, 0);
            adde(i+n, T, 1, 0);
            for(int j = 1; j <= n; j++) {
                adde(i, j+n, inf, -w[i][j]);
            }
        }
        cout << -MinCostFlow(S, T, N) << endl;
    }
    return 0;
}