網絡流訓練好題。。。但是要給差評！蒟蒻表示這就是板子題，然后做了半個小時T T

先跑一邊最大流，得到第一問答案ans。

第二問：原先的邊不動，費用為0。

然后對每條邊在上面再加一條邊，流量為inf，費用為修改費用。

n向T連一條邊，流量為ans + k，費用為0。

跑一邊費用流即可。

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1 /************************************************************** 2 Problem: 1834 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:64 ms 7 Memory:1296 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 const int N = 1005; 16 const int M = 10005; 17 const int inf = (int) 1e9; 18 19 struct Edges { 20 int x, y, c, w; 21 } E[M]; 22 23 struct edges { 24 int next, to, f, cost; 25 edges() {} 26 edges(int _n, int _t, int _f, int _c) : next(_n), to(_t), f(_f), cost(_c) {} 27 } e[M << 1]; 28 29 int n, m, k, last_ans; 30 int S, T; 31 int first[N], tot; 32 int q[N], d[N], g[N]; 33 bool v[N]; 34 35 inline int read() { 36 int x = 0, sgn = 1; 37 char ch = getchar(); 38 while (ch < '0' || '9' < ch) { 39 if (ch == '-') sgn = -1; 40 ch = getchar(); 41 } 42 while ('0' <= ch && ch <= '9') { 43 x = x * 10 + ch - '0'; 44 ch = getchar(); 45 } 46 return sgn * x; 47 } 48 49 inline void Add_Edges(int x, int y, int f, int c) { 50 e[++tot] = edges(first[x], y, f, c), first[x] = tot; 51 e[++tot] = edges(first[y], x, 0, -c), first[y] = tot; 52 } 53 54 bool bfs() { 55 memset(d, 0, sizeof(d)); 56 q[1] = S, d[S] = 1; 57 int l, r, x, y; 58 for (l = r = 1; l != (r + 1) % N; (++l) %= N) 59 for (x = first[q[l]]; x; x = e[x].next){ 60 y = e[x].to; 61 if (!d[y] && e[x].f) 62 q[(++r) %= N] = y, d[y] = d[q[l]] + 1; 63 } 64 65 return d[T]; 66 } 67 68 int dinic(int p, int limit) { 69 if (p == T || !limit) return limit; 70 int x, y, tmp, rest = limit; 71 for (x = first[p]; x; x = e[x].next) 72 if (d[y = e[x].to] == d[p] + 1 && e[x].f && rest) { 73 tmp = dinic(y, min(rest, e[x].f)); 74 rest -= tmp; 75 e[x].f -= tmp, e[x ^ 1].f += tmp; 76 if (!rest) return limit; 77 } 78 if (limit == rest) d[p] = 0; 79 return limit - rest; 80 } 81 82 int Dinic() { 83 int res = 0, x; 84 while (bfs()) 85 res += dinic(S, inf); 86 return res; 87 } 88 89 void work1() { 90 int i; 91 memset(first, 0, sizeof(first)); 92 for (i = tot = 1; i <= m; ++i) 93 Add_Edges(E[i].x, E[i].y, E[i].c, 0); 94 S = 1, T = n; 95 printf("%d ", last_ans = Dinic()); 96 } 97 98 inline int calc() { 99 int flow = inf, x; 100 for (x = g[T]; x; x = g[e[x ^ 1].to]) 101 flow = min(flow, e[x].f); 102 for (x = g[T]; x; x = g[e[x ^ 1].to]) 103 e[x].f -= flow, e[x ^ 1].f += flow; 104 return flow; 105 } 106 107 bool spfa() { 108 int x, y, l, r; 109 for (x = 1; x <= T; ++x) 110 d[x] = inf; 111 d[S] = 0, v[S] = 1, q[0] = S; 112 for(l = r = 0; l != (r + 1) % N; ++l %= N) { 113 for (x = first[q[l]]; x; x = e[x].next) 114 if (d[q[l]] + e[x].cost < d[y = e[x].to] && e[x].f) { 115 d[y] = d[q[l]] + e[x].cost; 116 g[y] = x; 117 if (!v[y]) 118 q[(++r) %= N] = y, v[y] = 1; 119 } 120 v[q[l]] = 0; 121 } 122 return d[T] != inf; 123 } 124 125 inline int work() { 126 int res = 0; 127 while (spfa()) 128 res += calc() * d[T]; 129 return res; 130 } 131 132 void work2() { 133 int i; 134 memset(first, 0, sizeof(first)); 135 for (i = tot = 1; i <= m; ++i) { 136 Add_Edges(E[i].x, E[i].y, E[i].c, 0); 137 Add_Edges(E[i].x, E[i].y, inf, E[i].w); 138 } 139 Add_Edges(n, n + 1, last_ans + k, 0); 140 S = 1, T = n + 1; 141 printf("%d\n", work()); 142 } 143 144 int main() { 145 int i; 146 n = read(), m = read(), k = read(); 147 for (i = 1; i <= m; ++i) 148 E[i].x = read(), E[i].y = read(), E[i].c = read(), E[i].w = read(); 149 work1(); 150 work2(); 151 return 0; 152 } View Code

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轉載于:https://www.cnblogs.com/rausen/p/4162739.html

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