POJ 3469 Dual Core CPU

时间：2014-09-25 16:43:29 阅读：508 评论：0 收藏：0 [点我收藏+]

标签：style blog http color io os java ar for

Dual Core CPU

Time Limit: 15000ms

Memory Limit: 131072KB

This problem will be judged on PKU. Original ID: 3469
64-bit integer IO format: %lld Java class name: Main

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, Aiand 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

Sample Output

Source

POJ Monthly--2007.11.25

解题：比较经典的最小割模型。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 20100;
18 struct arc{
19     int to,flow,next;
20     arc(int x = 0,int y = 0,int z = -1){
21         to = x;
22         flow = y;
23         next = z;
24     }
25 };
26 arc e[1000000];
27 int head[maxn],d[maxn],cur[maxn],tot,S,T;
28 int q[maxn<<2],hd,tl;
29 void add(int u,int v,int w){
30     e[tot] = arc(v,w,head[u]);
31     head[u] = tot++;
32     e[tot] = arc(u,0,head[v]);
33     head[v] = tot++;
34 }
35 bool bfs(){
36     memset(d,-1,sizeof(d));
37     hd = tl = 0;
38     q[tl++] = S;
39     d[S] = 1;
40     while(hd < tl){
41         int u = q[hd++];
42         for(int i = head[u]; ~i; i = e[i].next){
43             if(e[i].flow && d[e[i].to] == -1){
44                 d[e[i].to] = d[u] + 1;
45                 q[tl++] = e[i].to;
46             }
47         }
48     }
49     return d[T] > -1;
50 }
51 int dfs(int u,int low){
52     if(u == T) return low;
53     int tmp = 0,a;
54     for(int &i = cur[u]; ~i; i = e[i].next){
55         if(e[i].flow && d[e[i].to] == d[u] + 1 && (a = dfs(e[i].to,min(e[i].flow,low)))){
56             tmp += a;
57             low -= a;
58             e[i].flow -= a;
59             e[i^1].flow += a;
60             if(!low) break;
61         }
62     }
63     if(!tmp) d[u] = -1;
64     return tmp;
65 }
66 int main() {
67     int u,v,w,ans,n,m;
68     while(~scanf("%d %d",&n,&m)){
69         memset(head,-1,sizeof(head));
70         ans = S = tot = 0;
71         T = n + 1;
72         for(int i = 1; i <= n; i++){
73             scanf("%d %d",&u,&v);
74             add(S,i,u);
75             add(i,T,v);
76         }
77         for(int i = 0; i < m; i++){
78             scanf("%d %d %d",&u,&v,&w);
79             add(u,v,w);
80             add(v,u,w);
81         }
82         while(bfs()){
83             memcpy(cur,head,sizeof(head));
84             ans += dfs(S,INF);
85         }
86         printf("%d\n",ans);
87     }
88     return 0;
89 }

View Code

加输入挂的ISAP在g++下表现不错

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <climits>
  7 #include <vector>
  8 #include <queue>
  9 #include <cstdlib>
 10 #include <string>
 11 #include <set>
 12 #include <stack>
 13 #define LL long long
 14 #define pii pair<int,int>
 15 #define INF 0x3f3f3f3f
 16 using namespace std;
 17 const int maxn = 20100;
 18 struct arc {
 19     int to,flow,next;
 20     arc(int x = 0,int y = 0,int z = -1) {
 21         to = x;
 22         flow = y;
 23         next = z;
 24     }
 25 };
 26 arc e[1000000];
 27 int head[maxn],gap[maxn],d[maxn],cur[maxn];
 28 int S,T,tot,hd,tl,q[maxn],p[maxn];
 29 int getnum() {
 30     char ch;
 31     int num = 0,flag = 0;
 32     while(((ch = getchar()) < ‘0‘ || ch > ‘9‘) && ch != ‘-‘)
 33         if(ch == EOF)  exit(0);
 34     if(ch == ‘-‘)   flag = 1;
 35     else num = ch - ‘0‘;
 36     while((ch = getchar()) >= ‘0‘ && ch <= ‘9‘)
 37         num = num * 10 + ch - ‘0‘;
 38     if(flag)   num *= -1;
 39     return num;
 40 }
 41 void add(int u,int v,int w) {
 42     e[tot] = arc(v,w,head[u]);
 43     head[u] = tot++;
 44     e[tot] = arc(u,0,head[v]);
 45     head[v] = tot++;
 46 }
 47 void bfs() {
 48     hd = tl = 0;
 49     memset(d,-1,sizeof(d));
 50     memset(gap,0,sizeof(gap));
 51     q[tl++] = T;
 52     d[T] = 0;
 53     while(hd < tl) {
 54         int u = q[hd++];
 55         ++gap[d[u]];
 56         for(int i = head[u]; ~i; i = e[i].next) {
 57             if(d[e[i].to] == -1) {
 58                 d[e[i].to] = d[u] + 1;
 59                 q[tl++] = e[i].to;
 60             }
 61         }
 62     }
 63 }
 64 int isap() {
 65     int maxFlow = 0,flow = INF,u = S;
 66     memcpy(cur,head,sizeof(head));
 67     bfs();
 68     while(d[S] < T) {
 69         int &i = cur[u];
 70         for(; ~i; i = e[i].next)
 71             if(e[i].flow && d[e[i].to] + 1 == d[u]) break;
 72         if(i > -1) {
 73             flow = min(flow,e[i].flow);
 74             p[u = e[i].to] = i;
 75             if(u == T) {
 76                 do {
 77                     int v = p[u];
 78                     e[v].flow -= flow;
 79                     e[v^1].flow += flow;
 80                     u = e[v^1].to;
 81                 } while(u != S);
 82                 maxFlow += flow;
 83                 flow = INF;
 84             }
 85         } else {
 86             if(--gap[d[u]] == 0) break;
 87             d[u] = T;
 88             cur[u] = head[u];
 89             for(int k = head[u]; ~k; k = e[k].next)
 90                 if(e[k].flow && d[e[k].to] + 1 < d[u])
 91                     d[u] = d[e[k].to] + 1;
 92                 ++gap[d[u]];
 93                 if(u != S) u = e[p[u]^1].to;
 94         }
 95     }
 96     return maxFlow;
 97 }
 98 int main() {
 99     int n,m,u,v,w;
100     while(~scanf("%d %d",&n,&m)) {
101         memset(head,-1,sizeof(head));
102         S = tot = 0;
103         T = n + 1;
104         for(int i = 1; i <= n; i++) {
105             //scanf("%d %d",&u,&v);
106             u = getnum();
107             v = getnum();
108             add(S,i,u);
109             add(i,T,v);
110         }
111         for(int i = 0; i < m; i++) {
112             //scanf("%d %d %d",&u,&v,&w);
113             u = getnum();
114             v = getnum();
115             w = getnum();
116             add(u,v,w);
117             add(v,u,w);
118         }
119         printf("%d\n",isap());
120     }
121     return 0;
122 }

View Code

POJ 3469 Dual Core CPU

标签：style blog http color io os java ar for

原文地址：http://www.cnblogs.com/crackpotisback/p/3992689.html

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