hdu 3667 Transportation 費用流+拆邊
阿新 • • 發佈:2019-01-29
題意:給出一張n個點m條邊的圖,從起點1送k個單位的貨物到終點n,其中每條邊有容量限制Ci,且每條邊的花費與貨物經過這條邊的大小成ai 的係數關係,即運輸x的貨物,要ai * x * x的花費,求是否能夠送到終點,且求最小花費。
由於c才為5,很容易想到將其拆邊,假設容量為3,則能依次得到容量大小為1,花費為1, 3, 5的三條邊(1+5+3 = 3*3),然後跑一發費用流即可。
#include <bits/stdc++.h> #include <map> #include <set> #include <queue> #include <stack> #include <cmath> #include <time.h> #include <vector> #include <cstdio> #include <string> #include <iomanip> ///cout << fixed << setprecision(13) << (double) x << endl; #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 #define ls rt << 1 #define rs rt << 1 | 1 #define pi acos(-1.0) #define eps 1e-8 #define Mp(a, b) make_pair(a, b) #define asd puts("asdasdasdasdasdf"); typedef long long ll; typedef pair <int, int> pl; //typedef __int64 LL; const int inf = 0x3f3f3f3f; const int N = 111, M = 500050; struct node{ int v, w, c, nxt; //c--->cap w--->cost(dis) }e[M]; int cnt, n, m, k; int head[N]; int dis[N]; //費用看成距離 int cap[N]; //流量 int cur[N]; //當前弧 int vis[N]; queue <int> q; int flow, cost; int st, ed; void init() { st = 0, ed = n+1; cnt = 0; memset( head, -1, sizeof( head ) ); } void add( int u, int v, int c, int w ) { e[cnt].v = v; e[cnt].c = c; e[cnt].w = w; e[cnt].nxt = head[u]; head[u] = cnt++; e[cnt].v = u; e[cnt].c = 0; e[cnt].w = -w; e[cnt].nxt = head[v]; head[v] = cnt++; } bool spfa() { memset( dis, inf, sizeof( dis ) ); memset( vis, 0, sizeof( vis ) ); while( !q.empty() ) q.pop(); cap[st] = inf; cur[st] = -1; dis[st] = 0; q.push( st ); while( !q.empty() ) { int u = q.front(); q.pop(); vis[u] = 0; for( int i = head[u]; ~i; i = e[i].nxt ) { int v = e[i].v, c = e[i].c, w = e[i].w; if( c && dis[v] > dis[u] + w ) { dis[v] = dis[u] + w; cap[v] = min( c, cap[u] ); cur[v] = i; if( !vis[v] ) { vis[v] = 1; q.push(v); } } } } if( dis[ed] == inf ) return 0; flow += cap[ed]; cost += cap[ed] * dis[ed]; for( int i = cur[ed]; ~i; i = cur[e[i^1].v] ) { e[i].c -= cap[ed]; e[i^1].c += cap[ed]; } return 1; } int MCMF() { flow = cost = 0; while( spfa() ); return cost; } int main() { while( ~scanf("%d%d%d", &n, &m, &k) ) { init(); for( int i = 1; i <= m; ++i ) { int u, v, a, c; scanf("%d%d%d%d", &u, &v, &a, &c); for( int j = 1; j <= c; ++j ) { add( u, v, 1, a * ( 2 * j - 1 ) ); } } add( st, 1, k, 0 ); add( n, ed, inf, 0 ); int ans = MCMF(); if( flow < k ) puts("-1"); else printf("%d\n", ans); } }