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Language: Default Dual Core CPU
Description As more and more computers are equipped with dual core CPU, SetagLilb, the Chief Technology Officer of TinySoft Corporation, decided to update their famous product - SWODNIW. The routine consists of N modules, and each of them should run in a certain core. The costs for all the routines to execute on two cores has been estimated. Let‘s define them as Ai and Bi. Meanwhile, M pairs Input There are two integers in the first line of input data, N and M (1 ≤ N ≤ 20000, 1 ≤ M ≤ 200000) . The next N lines, each contains two integer, Ai and Bi. In the following M lines, each contains three integers: a, b, w. The meaning is that if module a and module b don‘t execute on the same core, you should pay extra w dollars for the data-exchange Output Output only one integer, the minimum total cost. Sample Input 3 1 1 10 2 10 10 3 2 3 1000 Sample Output 13 Source POJ Monthly--2007.11.25, Zhou Dong |
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题意：现在有n个模块，两个CPU A和B，每个模块要么在A上运行，要么在B上运行，给出每个模块在A和B机器上运行所需要的费用。接着m行，每行 a，b，w三个数字。表示如果a模块和b模块不在同一个机器上运行的话，需要额外花费w来共享数据。现在要求出运行所有任务最小的花费是多少。
思路：将两个CPU视为源点和汇点，对第i个模块在每个CPU中的耗费Ai和Bi，从源点向顶点i连接一条容量为Ai的弧，从顶点i向汇点连接一条容量为Bi的弧；对于a模块和b模块在不同CPU中运行造成的耗费w，从顶点a向b连容量为w的双向边。求最小割即求最大流。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 20005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
const int MAXM = 480010;
typedef long long ll;
using namespace std;
struct Edge
{
int to,next,cap,flow;
}edge[MAXM];
int n,m,s,t;
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
tol=0;
memset(head,-1,sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u,int v,int w,int rw=0)
{
edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u];
edge[tol].flow=0; head[u]=tol++;
edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v];
edge[tol].flow=0; head[v]=tol++;
}
//输入参数：起点，终点，点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start,int end,int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u=start;
pre[u]=-1;
gap[0]=N;
int ans=0;
while (dep[start]<N)
{
if (u==end)
{
int Min=INF;
for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])
if (Min>edge[i].cap-edge[i].flow)
Min=edge[i].cap-edge[i].flow;
for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])
{
edge[i].flow+=Min;
edge[i^1].flow-=Min;
}
u=start;
ans+=Min;
continue;
}
bool flag=false;
int v;
for (int i=cur[u];i!=-1;i=edge[i].next)
{
v=edge[i].to;
if (edge[i].cap-edge[i].flow && dep[v]+1==dep[u])
{
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if (flag)
{
u=v;
continue;
}
int Min=N;
for (int i=head[u];i!=-1;i=edge[i].next)
if (edge[i].cap-edge[i].flow && dep[edge[i].to]<Min)
{
Min=dep[edge[i].to];
cur[u]=i;
}
gap[dep[u]]--;
if (!gap[dep[u]]) return ans;
dep[u]=Min+1;
gap[dep[u]]++;
if (u!=start) u=edge[pre[u]^1].to;
}
return ans;
}
int main()
{
int i,j;
while (~sff(n,m))
{
int a,b,c;
init();
s=0;t=n+1;
FRE(i,1,n)
{
sff(a,b);
addedge(s,i,a);
addedge(i,t,b);
}
FRE(i,1,m)
{
sfff(a,b,c);
addedge(a,b,c,c);
}
printf("%d\n",sap(s,t,t+1));
}
return 0;
}
/*
3 1
1 10
2 10
10 3
2 3 1000
*/