Dual Core CPU
| Time Limit: 15000MS |
|
Memory Limit: 131072K |
| Total Submissions: 12960 |
|
Accepted: 5553 |
| Case Time Limit: 5000MS |
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 of modules need to do some data-exchange. If they are running on the same core, then the cost of this action can be ignored. Otherwise, some extra cost are needed. You should arrange wisely to minimize the total cost.
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 between them.
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
这题点有20000，边有200000，所以要写链表或者数组模拟
终于找了个能看懂的了
而且我觉着这代码特别神有几个地方
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define nmax 20010
5#define emax 200010
6#define inf 1<<30
7int nn;
8int head[nmax];
9struct node
10
{
11
int v,next,w;
12
};
13struct node edge[emax*8];
14int cnt,n,m,s,t;
15void add(int u,int v,int w)
16{
17 edge[cnt].v=v;
18 edge[cnt].w=w;
19 edge[cnt].next=head[u];
20 head[u]=cnt++;
21 edge[cnt].v=u;
22 edge[cnt].w=0;
23 edge[cnt].next=head[v];
24 head[v]=cnt++;
25}
26int sap()
27{
28 int pre[nmax],cur[nmax],dis[nmax],gap[nmax];
29 int flow,aug,u;
30 int flag;
31 int i;
32 flow=0;
33 aug=inf;
34 for(i=0; i<=nn; i++)
35
{
36 cur[i]=head[i];
37 gap[i]=0;
38 dis[i]=0;
39
}
40 gap[s]=nn;
41 u=s;
42 pre[s]=s;
43 while(dis[s]<nn)
44 {
45 flag=0;
46 for(int &j=cur[u]; j!=-1; j=edge[j].next)
47 {
48 int v=edge[j].v;
49 if(edge[j].w>0&&dis[u]==dis[v]+1)
50 {
51 flag=1;
52 if(edge[j].w<aug)aug=edge[j].w;
53 pre[v]=u;
54 u=v;
55 if (u==t)
56 {
57 flow+=aug;
58 while(u!=s)
59 {
60 u=pre[u];
61 edge[cur[u]].w-=aug;
62 edge[cur[u]^1].w+=aug;
63 //why?解释偶数异或1为偶数+1，奇数异或1为奇数-1，
64 //显然我们存的边是从0开始存的，
65 //所以偶数,偶数+1是残量网格中的两条边(无向边)
66 }
67 aug=inf;
68 }
69 break;
70 }
71 }
72 if (flag) continue;
73 int mindis=nn;
74 for(int j=head[u];j!=-1;j=edge[j].next)
75 {
76 int v=edge[j].v;
77 if (edge[j].w>0&&dis[v]<mindis)
78 {
79 mindis=dis[v];
80 cur[u]=j;
81 }
82 }
83 if (--gap[dis[u]]==0)//间隙优化
84 {
85 break;
86 }
87 dis[u]=mindis+1;
88 gap[dis[u]]++;
89 u=pre[u];
90 }
91 return flow;
92}
93int main()
94{
95 int a,b,c,i;
96 int ans;
97 while (scanf("%d%d",&n,&m)!=EOF)
98 {
99 s=0;
100 t=n+1;
101 cnt=0;
102 memset(head,-1,sizeof(head));
103 for(i=1; i<=n; i++)
104 {
105 scanf("%d%d",&a,&b);
106 add(s,i,a);
107 add(i,t,b);
108 }
109 for(i=1; i<=m; i++)
110 {
111 scanf("%d%d%d",&a,&b,&c);
112 add(a,b,c);
113 add(b,a,c);
114 }
115 ans=0;
116 nn=n+2;
117 ans=sap();
118 printf("%d\n",ans);
119 }
120 return 0;
121}
122
有两个需要特别注意的循环撒
那个int j那个很奇怪，怪我语言没学好