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【CSUOJ 1623】Inspectors

2015-05-18 15:14 267 查看
对每一个点进行拆点,并且每个入点和出点的容量设为1,因为每个点的出度与入度都为1且不存在自己连自己的边,故最后得到的图都为环。跑最小费用流就好了e

代码如下:

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 222;
const int M = 11111;
const int INF = 1e9;
struct edge{
int to,next;
int cap,flow,cost;
}e[M << 1];
int head
,sz,res;
bool inq
;
int p
,d
,a
;
void addedge(int u,int v,int cap,int cost){
e[sz].to = v;e[sz].next = head[u];
e[sz].cap = cap;e[sz].flow = 0;e[sz].cost = cost;
head[u] = sz ++;
e[sz].to = u;e[sz].next = head[v];
e[sz].cap = 0;e[sz].flow = 0;e[sz].cost = -cost;
head[v] = sz ++;
}
void init(){
memset(head,-1,sizeof(head));
sz = 0;
}
int SPFA(int s,int t){
queue<int> Q;
memset(inq,0,sizeof(inq));
for(int i = 1 ; i <= t ; i ++) d[i] = INF;
inq[s] = 1;d[s] = 0;p[s] = -1;
a[s] = INF;
Q.push(s);
while(!Q.empty()){
int u = Q.front();Q.pop();
inq[u] = 0;
for(int i = head[u] ; i != -1 ; i = e[i].next){
int v = e[i].to;
if(e[i].cap - e[i].flow > 0 && e[i].cost + d[u] < d[v]){
d[v] = d[u] + e[i].cost;
a[v] = min(e[i].cap - e[i].flow,a[u]);
p[v] = i;
if(!inq[v]){
inq[v] = 1;
Q.push(v);
}
}
}
}
int u = t;
if(d[t] == INF) return 0;
res += d[t] * a[t];
while(u != s){
e[ p[u] ].flow += a[t];
e[ p[u] ^ 1 ].flow -= a[t];
u = e[ p[u] ^ 1 ].to;
}
return 1;
}
int txt = 1;
void MCMF(int s,int t){
res = 0;
while(SPFA(s,t));
printf("Case %d: %d\n",txt ++,res);
}
void solve(){
init();
int s,t,c,n;
scanf("%d",&n);
s = 2 * n + 1;t = 2 * n + 2;
for(int i = 1 ; i <= n ; i ++){
addedge(s,i,1,0);
addedge(i + n,t,1,0);
}
for(int i = 1 ; i < n ; i ++){
for(int j = i + 1 ; j <= n ; j ++){
scanf("%d",&c);
addedge(i,j + n,1,c);
addedge(j,i + n,1,c);
}
}
MCMF(s,t);
}
int main()
{
int _;
scanf("%d",&_);
while(_--) solve();
return 0;
}
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