hdu 3667 Transportation 费用流+拆边
2015-09-23 15:00
363 查看
题意:给出一张n个点m条边的图,从起点1送k个单位的货物到终点n,其中每条边有容量限制Ci,且每条边的花费与货物经过这条边的大小成ai 的系数关系,即运输x的货物,要ai * x * x的花费,求是否能够送到终点,且求最小花费。
由于c才为5,很容易想到将其拆边,假设容量为3,则能依次得到容量大小为1,花费为1, 3, 5的三条边(1+5+3 = 3*3),然后跑一发费用流即可。
由于c才为5,很容易想到将其拆边,假设容量为3,则能依次得到容量大小为1,花费为1, 3, 5的三条边(1+5+3 = 3*3),然后跑一发费用流即可。
#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 ; int dis ; //费用看成距离 int cap ; //流量 int cur ; //当前弧 int vis ; 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); } }
相关文章推荐
- android studio引入最新版银联支付功能
- java类加载器总结
- hexagon绝对路径
- 脑子里冒出了几个公式
- git移植操作
- 学习Oracle数据库入门到精通教程资料合集
- 本机Ajax异步通信
- nginx 并发数问题思考:worker_connections,worker_processes与 max clients
- HTML-右键菜单
- Java编程:比对两个文本文件,标记相同和不同之处
- 优化数据页面(32)——显示在一屏内
- VS2010安装MSDN(转载)
- KVM迁移
- css之定位元素
- loadrunner解决“服务器正在运行中”方法
- 对于java中clone()函数的理解
- 装双系统(win7/win8/ubuntu)问题总结
- JS小数位保留两位小数--toFixed()
- Raw_Socket原始套接字
- HighCharts参数配置