洛谷P2050 美食节

修车加强版。发现每个厨师拆成p个点太浪费了，毕竟总共用到的才p个点。于是从下往上一个一个加，加到满流就停。

论动态加点费用流的正确姿势......

我自己加总是出现负环...我是每次加一整层，然后跑完这一层再加下一层，这样会显而易见的出现负环......

然后我们发现如果每增广一流量就加边就不会出现这种毒瘤现象，因为每次加的一定比增广的劣......

注意一定要动态开点，不能只用一个点代表厨师。否则可能出现厨师的第一次给了多个菜的情况...

#include <bits/stdc++.h>

const int N = 100010, INF = 0x7f7f7f7f;

struct Edge {
    int nex, v, c, len;
    Edge(int Nex = 0, int V = 0, int C = 0, int Len = 0) {
        nex = Nex;
        v = V;
        c = C;
        len = Len;
    }
}edge[5000010]; int tp = 1;

int e[N], lm, now[N], pre[N], flow[N], p[N], cnt[N], d[N];
int val[110][110];
bool vis[N];
std::queue<int> Q;

inline void add(int x, int y, int z, int w) {
    edge[++tp] = Edge(e[x], y, z, w);
    e[x] = tp;
    edge[++tp] = Edge(e[y], x, 0, -w);
    e[y] = tp;
    return;
}

inline bool SPFA(int s, int t) {
    memset(d + 1, 0x7f, lm * sizeof(int));
    vis[s] = 1;
    flow[s] = INF;
    d[s] = 0;
    Q.push(s);
    while(!Q.empty()) {
        int x = Q.front();
        Q.pop();
        vis[x] = 0;
        for(int i = e[x]; i; i = edge[i].nex) {
            int y = edge[i].v;
            if(d[y] > d[x] + edge[i].len && edge[i].c) {
                d[y] = d[x] + edge[i].len;
                //printf("%d -> %d 
", x, y);
                flow[y] = std::min(flow[x], edge[i].c);
                pre[y] = i;
                if(!vis[y]) {
                    vis[y] = 1;
                    Q.push(y);
                }
            }
        }
    }
    return d[t] < INF;
}

inline void update(int s, int t) {
    int f = flow[t];
    while(t != s) {
        int i = pre[t];
        edge[i].c -= f;
        edge[i ^ 1].c += f;
        t = edge[i ^ 1].v;
    }
    return;
}

int main() {

    int n, m, tot = 0;
    scanf("%d%d", &n, &m);
    lm = n;
    int s = ++lm;
    int t = ++lm;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &p[i]);
        add(s, i, p[i], 0);
        tot += p[i];
    }
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            scanf("%d", &val[i][j]);
        }
    }
    /// sol
    for(int i = 1; i <= m; i++) {
        add(++lm, t, 1, 0);
        cnt[i] = 1;
        now[i] = tp - 1;
        for(int j = 1; j <= n; j++) {
            add(j, lm, 1, val[j][i]);
        }
    }

    int ans = 0;
    while(tot) {
        //printf("tot = %d ans = %d 
", tot, ans);
        SPFA(s, t);
        ans += d[t] * flow[t];
        tot -= flow[t];
        update(s, t);
        for(int j = 1; j <= m; j++) {
            if(edge[now[j]].c) continue;
            add(++lm, t, 1, 0);
            now[j] = tp - 1;
            cnt[j]++;
            for(int i = 1; i <= n; i++) {
                add(i, lm, 1, cnt[j] * val[i][j]);
            }
            break;
        }
    }

    printf("%d
", ans);
    return 0;
}