bzoj1070 修车&& bzoj2879美食节 【费用流】

bzoj1070:

　　把每个工人拆成汽车那么多个点，假如说 工人(i, j) 和 汽车k 连边，那就代表第i个工人倒数第j个修汽车k，那么这条边对以后的贡献就是k*time[i修k]。

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define mp make_pair
 #define pb push_back
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second
 using namespace std;
 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 //*********************************

 const int maxn = , maxm = ;

 struct Ed {
     int u, v, nx, c, w; Ed() {}
     Ed(int _u, int _v, int _nx, int _c, int _w) :
         u(_u), v(_v), nx(_nx), c(_c), w(_w) {}
 } E[maxm << ];
 int G[maxn], edtot;
 void addedge(int u, int v, int c, int w) {
     E[edtot] = (Ed){u, v, G[u], c, w};
     G[u] = edtot++;
     E[edtot] = (Ed){v, u, G[v], , -w};
     G[v] = edtot++;
 }

 bool vis[maxn]; int dis[maxn], s, t;
 bool spfa() {
     static int que[maxm]; int qh(), qt();
     rep(i, s, t) vis[i] = , dis[i] = 0x3f3f3f3f;
     dis[que[++qt] = s] = ; vis[s] = ;
     while (qh != qt) {
         int x = que[++qh]; vis[x] = ;
         for (int i = G[x]; i != -; i = E[i].nx) {
             if (E[i].c && dis[E[i].v] > dis[x] + E[i].w) {
                 dis[E[i].v] = dis[x] + E[i].w;
                 if (!vis[E[i].v]) vis[que[++qt] = E[i].v] = ;
             }
         }
     }
     return dis[t] != 0x3f3f3f3f;
 }
 int ans, cur[maxn];
 int dfs(int u, int rm) {
     vis[u] = ;
     if (u == t) return rm;
     int rm1 = rm;
     for (int &i = cur[u]; i != -; i = E[i].nx) {
         if (E[i].c && !vis[E[i].v] && dis[E[i].v] == dis[u] + E[i].w) {
             int flow = dfs(E[i].v, min(rm, E[i].c));
             E[i].c -= flow, E[i ^ ].c += flow;
             ans += flow * E[i].w;
             if ((rm -= flow) == ) break;
         }
     }
     if (rm1 == rm) dis[u] = ;
     return rm1 - rm;
 }

 int a[][];
 int main() {
     int m, n; scanf("%d%d", &m, &n);
     rep(i, , n) rep(j, , m) scanf("%d", &a[i][j]);
     s = , t = n + n * m + ;
     memset(G, -, sizeof(G));
     rep(i, , n) addedge(s, i, , );
     rep(i, n + , n + n * m) addedge(i, t, , );
     rep(i, , n) {
         rep(j, , m) {
             rep(k, , n) {
                 addedge(i, n + (j - ) * n + k, , k * a[i][j]);
             }
         }
     }
     while (spfa()) memcpy(cur, G, sizeof(G)),dfs(s, 0x3f3f3f3f);
     printf("%.2lf\n", 1.0 * ans / n);
     return ;
 }

bzoj2879:

　　首先把每个食物和厨师连边，一开始只用和每个厨师的倒数第一这个时间段连边，如果用了的话再用这个厨师的倒数第二去连边。

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define pb push_back
 #define mp make_pair
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second
 using namespace std;
 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 //**********************************

 const int maxn = , maxm = ;

 struct Ed {
     int u, v, nx, c, w; Ed() {}
     Ed(int _u, int _v, int _nx, int _c, int _w) :
         u(_u), v(_v), nx(_nx), c(_c), w(_w) {}
 } E[maxm];
 int G[maxn], edtot = ;
 void addedge(int u, int v, int c, int w) {
     E[++edtot] = Ed(u, v, G[u], c, w);
     G[u] = edtot;
     E[++edtot] = Ed(v, u, G[v], , -w);
     G[v] = edtot;
 }

 int tot, n, m;

 bool vis[maxn]; int dis[maxn], s, t, pre[maxn];
 bool spfa() {
     static int que[maxn]; int qh(), qt();
     rep(i, s, t) vis[i] = , dis[i] = inf;
     vis[que[++qt] = s] = , dis[s] = ;
     while (qh != qt) {
         int x = que[++qh]; if (qh == t) qh = ;
         for (int i = G[x]; i; i = E[i].nx) {
             if (E[i].c && dis[E[i].v] > dis[x] + E[i].w) {
                 dis[E[i].v] = dis[x] + E[i].w;
                 pre[E[i].v] = i;
                 if (!vis[E[i].v]) {
                     vis[que[++qt] = E[i].v] = ;
                     if (qt == t) qt = ;
                 }
             }
         }
         vis[x] = ;
     }
     return dis[t] != inf;
 }
 int a[][];
 int ans;
 void mcf() {
     int flow = inf, x, y;
     for (int i = pre[t]; i; i = pre[E[i].u]) {
         flow = min(flow, E[i].c);
         if (E[i].v == t) {
             x = (E[i].u - ) / tot + ; y = E[i].u % tot + ;
         }
     }
     for (int i = pre[t]; i; i = pre[E[i].u]) {
         E[i].c -= flow, E[i ^ ].c += flow, ans += flow * E[i].w;
     }
     addedge((x - ) * tot + y, t, , );
     for (int i = ; i <= n; i++)
         addedge(m * tot + i, (x - ) * tot + y, , y * a[i][x]);
 }

 int main() {
     scanf("%d%d", &n, &m);
     static int c[];
     rep(i, , n) scanf("%d", c + i), tot += c[i];
     rep(i, , n) rep(j, , m) scanf("%d", &a[i][j]);
     s = , t = m * tot + n + ;
     rep(i, , n) addedge(s, m * tot + i, c[i], );
     rep(i, , m) addedge((i - ) * tot + , t, , );
     rep(i, , m) rep(j, , n) addedge(m * tot + j, (i - ) * tot + , , a[j][i]);
     while (spfa()) mcf();
     printf("%d\n", ans);
     return ;
 }

　　注意，把从食物向厨师连边比较快。