poj 2455 Secret Milking Machine 二分+最大流 sap

题目:p条路,连接n个节点,现在需要从节点1到节点n,不重复走过一条路且走t次,最小化这t次中连接两个节点最长的那条路的值。

分析:二分答案,对于<=二分的值的边建边,跑一次最大流即可。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; typedef long long ll;
typedef unsigned long long ull; #define debug puts("here")
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)
#define pb push_back
#define RD(n) scanf("%d",&n)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)
#define All(vec) vec.begin(),vec.end()
#define MP make_pair
#define PII pair<int,int>
#define PQ priority_queue
#define cmax(x,y) x = max(x,y)
#define cmin(x,y) x = min(x,y)
#define Clear(x) memset(x,0,sizeof(x))
/* #pragma comment(linker, "/STACK:1024000000,1024000000") int size = 256 << 20; // 256MB
char *p = (char*)malloc(size) + size;
__asm__("movl %0, %%esp\n" :: "r"(p) ); */ /******** program ********************/ const int MAXN = 1005;
const int MAXM = 100005;
const int INF = 1e9; int po[MAXN],tol;
int gap[MAXN],dis[MAXN],arc[MAXN],pre[MAXN],cur[MAXN];
int n,m,vs,vt,t; struct Edge{
int y,f,next;
}edge[MAXM]; struct node{
int x,y,l;
}p[MAXM]; void Add(int x,int y,int f){
edge[++tol].y = y;
edge[tol].f = f;
edge[tol].next = po[x];
po[x] = tol;
}
void add(int x,int y,int f){
Add(x,y,f);
Add(y,x,f); // 正边、反边流量均为f
} int sap(){
memset(dis,0,sizeof(dis));
memset(gap,0,sizeof(gap));
gap[0] = vt;
rep1(i,vt)
arc[i] = po[i]; int ans = 0;
int aug = INF;
int x = vs; while(dis[vs]<vt){
bool ok = false;
cur[x] = aug;
for(int i=arc[x];i;i=edge[i].next){
int y = edge[i].y;
if(edge[i].f>0&&dis[y]+1==dis[x]){
ok = true;
pre[y] = arc[x] = i;
aug = min(aug,edge[i].f);
x = y;
if(x==vt){
ans += aug;
while(x!=vs){
edge[pre[x]].f -= aug;
edge[pre[x]^1].f += aug;
x = edge[pre[x]^1].y;
}
aug = INF;
}
break;
}
}
if(ok)
continue;
int MIN = vt-1;
for(int i=po[x];i;i=edge[i].next)
if(edge[i].f>0&&dis[edge[i].y]<MIN){
MIN = dis[edge[i].y];
arc[x] = i;
}
if(--gap[dis[x]]==0)
break;
dis[x] = ++ MIN;
++ gap[dis[x]];
if(x!=vs){
x = edge[pre[x]^1].y;
aug = cur[x];
}
}
return ans;
} inline bool ok(int mid){
Clear(po);
tol = 1; vs = 1;
vt = n; rep1(i,m)
if(p[i].l<=mid)
add(p[i].x,p[i].y,1); // 此处是无向边 return sap()>=t;
} int main(){ #ifndef ONLINE_JUDGE
freopen("sum.in","r",stdin);
//freopen("sum.out","w",stdout);
#endif while(~RD3(n,m,t)){
int l = 10000000 , r = 0;
rep1(i,m){
RD3(p[i].x,p[i].y,p[i].l);
cmin(l,p[i].l);
cmax(r,p[i].l);
} int ans = 0;
while(l<=r){
int mid = (l+r)>>1;
if(ok(mid)){
r = mid-1;
ans = mid;
}else
l = mid+1;
}
cout<<ans<<endl;
} return 0;
}

  

上一篇:SSM框架整合首只拦路虎——Eclipse新建Maven Project界面select an archetype 空白


下一篇:zookeeper集群的搭建以及hadoop ha的相关配置