POJ 3469 Dual Core CPU (最小割建模)
题意
现在有n个任务,两个机器A和B,每个任务要么在A上完成,要么在B上完成,而且知道每个任务在A和B机器上完成所需要的费用。然后再给m行,每行 a,b,w三个数字。表示如果a任务和b任务不在同一个机器上工作的话,需要额外花费w。现在要求出完成所有任务最小的花费是多少。
思路
上次做的构图题是基于割截断s->t流与题目联系的,而这道题的构图则是基于割把流网络的点划分成了S、T点集,并且不同点集间的边都是割边。这是目前我所接触到的最小割的建模类型。 回到本题构图:源点向任务连一条Ai容量的边,任务向汇点连一条Bi容量的边,如果两任务在不同Core上需要代价则连一条(i,j,cost)的无向边。
代码
#include
#include
#include
#include
#include
#include
#define MID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXV = 505;
const int MAXE = 20005;
const int oo = 0x3fffffff;
struct node{
int u, v, flow;
int opp;
int next;
};
struct Dinic{
node arc[MAXE];
int vn, en, head[MAXV]; //vn点个数(包括源点汇点),en边个数
int cur[MAXV]; //当前弧
int q[MAXV]; //bfs建层次图时的队列
int path[MAXE], top; //存dfs当前最短路径的栈
int dep[MAXV]; //各节点层次
void init(int n){
vn = n;
en = 0;
mem(head, -1);
}
void insert_flow(int u, int v, int flow){
arc[en].u = u;
arc[en].v = v;
arc[en].flow = flow;
arc[en].opp = en + 1;
arc[en].next = head[u];
head[u] = en ++;
arc[en].u = v;
arc[en].v = u;
arc[en].flow = 0; //反向弧
arc[en].opp = en - 1;
arc[en].next = head[v];
head[v] = en ++;
}
bool bfs(int s, int t){
mem(dep, -1);
int lq = 0, rq = 1;
dep[s] = 0;
q[lq] = s;
while(lq < rq){
int u = q[lq ++];
if (u == t){
return true;
}
for (int i = head[u]; i != -1; i = arc[i].next){
int v = arc[i].v;
if (dep[v] == -1 && arc[i].flow > 0){
dep[v] = dep[u] + 1;
q[rq ++] = v;
}
}
}
return false;
}
int solve(int s, int t){
int maxflow = 0;
while(bfs(s, t)){
int i, j;
for (i = 1; i <= vn; i ++) cur[i] = head[i];
for (i = s, top = 0;;){
if (i == t){
int mink;
int minflow = 0x3fffffff;
for (int k = 0; k < top; k ++)
if (minflow > arc[path[k]].flow){
minflow = arc[path[k]].flow;
mink = k;
}
for (int k = 0; k < top; k ++)
arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow;
maxflow += minflow;
top = mink; //arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内).
i = arc[path[top]].u;
}
for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){
int v = arc[j].v;
if (arc[j].flow && dep[v] == dep[i] + 1)
break;
}
if (j != -1){
path[top ++] = j;
i = arc[j].v;
}
else{
if (top == 0) break;
dep[i] = -1;
i = arc[path[-- top]].u;
}
}
}
return maxflow;
}
}dinic;
bool vis[MAXV];
bool reach(int u, int p){
vis[u] = 1;
if (u == p)
return true;
for (int i = dinic.head[u]; i != -1; i = dinic.arc[i].next){
if (i % 2 == 1) continue;
int v = dinic.arc[i].v;
if (vis[v] || dinic.arc[i].flow <= 0) continue;
if (reach(v, p))
return true;
}
return false;
}
int work(int n){
vector seg;
for (int i = 0; i < dinic.en; i += 2){
if (dinic.arc[i].flow == 0){
mem(vis, 0);
int u = dinic.arc[i].u;
int v = dinic.arc[i].v;
if (reach(n+1, u) && reach(v, n+2)){
seg.push_back(i/2+1);
}
}
}
for (int i = 0; i < (int)seg.size(); i ++){
if (i == 0)
printf("%d", seg[i]);
else
printf(" %d", seg[i]);
}
puts("");
}
int main(){
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
int n, m, l;
while(scanf("%d %d %d", &n, &m, &l), n){
dinic.init(n+m+2);
for (int i = 0; i < l; i ++){
int u,v,w;
scanf("%d %d %d", &u, &v, &w);
dinic.insert_flow(u==0?n+m+2:u, v==0?n+m+2:v, w);
}
for (int i = 1; i <= n; i ++){
dinic.insert_flow(n+m+1, i, oo);
}
dinic.solve(n+m+1, n+m+2);
work(n+m);
}
return 0;
}
举杯独醉,饮罢飞雪,茫然又一年岁。 ------AbandonZHANG