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POJ 3469 Dual Core CPU （求最小割）
POJ 3469 Dual Core CPU
链接：http://poj.org/problem?id=3469
题意：有两个cpu，n个模块。每个模块运行在连个cpu上运行时间不同。有m对模块之间要进行信息交互，如果模块在同一个cpu，那么进行信息交互时不需要额外时间，否则要花额外的时间。问怎么样分配模块，能够使得花费的时间最少。
思路：要将模块分给两个cpu，同时要使得时间最少，其实就是求最小割。那么题目可以转化为求最大流。
代码：
/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define LINF (1LL<<60)
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
//#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 121000;
const int maxm = 1410000;
int st, ed, n, m;
struct node {
int v; // vertex
int cap; // capacity
int flow; // current flow in this arc
int nxt;
} e[maxm * 2];
int g[maxn], cnt;
void add(int u, int v, int c) {
e[++cnt].v = v;
e[cnt].cap = c;
e[cnt].flow = 0;
e[cnt].nxt = g[u];
g[u] = cnt;
e[++cnt].v = u;
e[cnt].cap = 0;
e[cnt].flow = 0;
e[cnt].nxt = g[v];
g[v] = cnt;
}
void init() {
mem(g, 0);
cnt = 1;
st = 0;
ed = n + 1;
int u, v, c;
for (int i = 1; i <= n; i++) {
scanf("%d%d", &u, &v);
add(st, i, u);
add(i, ed, v);
}
while(m--) {
scanf("%d%d%d", &u, &v, &c);
add(u, v, c);
add(v, u, c);
}
n = n + 3;
}
int dist[maxn], numbs[maxn], q[maxn];
void rev_bfs() {
int font = 0, rear = 1;
for (int i = 0; i <= n; i++) { //n为总点数
dist[i] = maxn;
numbs[i] = 0;
}
q[font] = ed;
dist[ed] = 0;
numbs[0] = 1;
while(font != rear) {
int u = q[font++];
for (int i = g[u]; i; i = e[i].nxt) {
if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue;
dist[e[i].v] = dist[u] + 1;
++numbs[dist[e[i].v]];
q[rear++] = e[i].v;
}
}
}
int maxflow() {
rev_bfs();
int u, totalflow = 0;
int curg[maxn], revpath[maxn];
for(int i = 0; i <= n; ++i) curg[i] = g[i];
u = st;
while(dist[st] < n) {
if(u == ed) { // find an augmenting path
int augflow = INF;
for(int i = st; i != ed; i = e[curg[i]].v)
augflow = min(augflow, e[curg[i]].cap);
for(int i = st; i != ed; i = e[curg[i]].v) {
e[curg[i]].cap -= augflow;
e[curg[i] ^ 1].cap += augflow;
e[curg[i]].flow += augflow;
e[curg[i] ^ 1].flow -= augflow;
}
totalflow += augflow;
u = st;
}
int i;
for(i = curg[u]; i; i = e[i].nxt)
if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break;
if(i) { // find an admissible arc, then Advance
curg[u] = i;
revpath[e[i].v] = i ^ 1;
u = e[i].v;
} else { // no admissible arc, then relabel this vertex
if(0 == (--numbs[dist[u]])) break; // GAP cut, Important!
curg[u] = g[u];
int mindist = n;
for(int j = g[u]; j; j = e[j].nxt)
if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]);
dist[u] = mindist + 1;
++numbs[dist[u]];
if(u != st)
u = e[revpath[u]].v; // Backtrack
}
}
return totalflow;
}
int main () {
while(~scanf("%d%d", &n, &m)) {
init();
printf("%d\n", maxflow());
}
return 0;
}
POJ 3469 Dual Core CPU （求最小割）
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