poj3469 Dual Core CPU

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双核CPU还行...

题意:

给定n个任务 由a核和b核完成分别需要ai,bi花费

同时m个约束形如(x,y,w) 如果x,y不在一个核上就额外花费w 求最小花费

 

Solution:

一开始想的拆点...最后发现图不连通

其实这个每个点向两边分别连ai,bi就行 因为流取得就是路径上所有边的最小值

没错就是源汇分别表示A,B

求一个最小割表示最小花费

约束的话就连一个双向边

Code:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 #include<iostream>
 5 #include<algorithm>
 6 #define ms(a,b) memset(a,b,sizeof a)
 7 #define rep(i,a,n) for(int i = a;i <= n;i++)
 8 #define per(i,n,a) for(int i = n;i >= a;i--)
 9 #define inf 2147483647
10 using namespace std;
11 typedef long long ll;
12 typedef double D;
13 #define eps 1e-8
14 ll read() {
15     ll as = 0,fu = 1;
16     char c = getchar();
17     while(c < '0' || c > '9') {
18         if(c == '-') fu = -1;
19         c = getchar();
20     }
21     while(c >= '0' && c <= '9') {
22         as = as * 10 + c - '0';
23         c = getchar();
24     }
25     return as * fu;
26 }
27 const int N = 20005;
28 const int M = 900005;
29 //head
30 int s = N-2,t = N-1;
31 int head[N],nxt[M],mo[M],cst[M],cnt = 1;
32 void _add(int x,int y,int w) {
33     mo[++cnt] = y;
34     cst[cnt] = w;
35     nxt[cnt] = head[x];
36     head[x] = cnt;
37 }
38 void add(int x,int y,int w) {
39     if(x^y) _add(x,y,w),_add(y,x,0);
40 }
41 
42 int dep[N],cur[N];
43 bool bfs() {
44     queue<int> q;
45     memcpy(cur,head,sizeof cur);
46     ms(dep,0),q.push(s),dep[s] = 1;
47     while(!q.empty()) {
48         int x = q.front();
49         q.pop();
50         for(int i = head[x];i;i = nxt[i]) {
51             int sn = mo[i];
52             if(!dep[sn] && cst[i]) {
53                 dep[sn] = dep[x] + 1;
54                 q.push(sn);
55             }
56         }
57     }
58     return dep[t];
59 }
60 
61 int dfs(int x,int flow) {
62     if(x == t || flow == 0) return flow;
63     int res = 0;
64     for(int &i = cur[x];i;i = nxt[i]) {
65         int sn = mo[i];
66         if(dep[sn] == dep[x] + 1 && cst[i]) {
67             int d = dfs(sn,min(cst[i],flow - res));
68             if(d) {
69                 cst[i] -= d,cst[i^1] += d;
70                 res += d;
71                 if(res == flow) break;
72             }
73         }
74     }
75     if(res ^ flow) dep[x] = 0;
76     return res;
77 }
78 
79 int DINIC() {
80     int ans = 0;
81     while(bfs()) ans += dfs(s,inf);
82     return ans;
83 }
84 
85 int n,m;
86 
87 int main() {
88     n = read(),m = read();
89     rep(i,1,n) {
90         add(s,i,read());
91         add(i,t,read());
92     }
93     rep(i,1,m) {
94         int x = read(),y = read(),w = read();
95         add(x,y,w),add(y,x,w);
96     }
97     printf("%d\n",DINIC());
98     return 0;
99 }

 

posted @ 2018-11-24 14:23  白怀潇  阅读(227)  评论(0)  编辑  收藏  举报