POJ 2391 最大流 二分 拆點 floyd

Ombrophobic Bovines

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16756		Accepted: 3656

Description

FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to create a rain evacuation plan so that all the cows can get to shelter before the rain begins. Weather forecasting is not always correct, though. In order to minimize false alarms, they want to sound the siren as late as possible while still giving enough time for all the cows to get to some shelter. 

The farm has F (1 <= F <= 200) fields on which the cows graze. A set of P (1 <= P <= 1500) paths connects them. The paths are wide, so that any number of cows can traverse a path in either direction. 

Some of the farm's fields have rain shelters under which the cows can shield themselves. These shelters are of limited size, so a single shelter might not be able to hold all the cows. Fields are small compared to the paths and require no time for cows to traverse. 

Compute the minimum amount of time before rain starts that the siren must be sounded so that every cow can get to some shelter.

Input

* Line 1: Two space-separated integers: F and P 

* Lines 2..F+1: Two space-separated integers that describe a field. The first integer (range: 0..1000) is the number of cows in that field. The second integer (range: 0..1000) is the number of cows the shelter in that field can hold. Line i+1 describes field i. 

* Lines F+2..F+P+1: Three space-separated integers that describe a path. The first and second integers (both range 1..F) tell the fields connected by the path. The third integer (range: 1..1,000,000,000) is how long any cow takes to traverse it.

Output

* Line 1: The minimum amount of time required for all cows to get under a shelter, presuming they plan their routes optimally. If it not possible for the all the cows to get under a shelter, output "-1".

Sample Input

3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120

Sample Output

110

Hint

OUTPUT DETAILS: 

In 110 time units, two cows from field 1 can get under the shelter in that field, four cows from field 1 can get under the shelter in field 2, and one cow can get to field 3 and join the cows from that field under the shelter in field 3. Although there are other plans that will get all the cows under a shelter, none will do it in fewer than 110 time units.

題意是有F個牛棚。裏面現在有多少隻牛 但是最多能容納多少隻牛  所以其他牛需要去別的棚子

棚子和棚子之間是無向圖。。。用floyd一遍就搞定了

這道題。。不知道怎麼回事。。無腦RE 重寫到第三遍的時候。才AC 的。。BUT 。。TELL MY WHY。。。

#include <iostream>  
#include <algorithm>  
#include <cstring>  
#include <string>  
#include <cstdio>  
#include <cmath>    
  
using namespace std;  
const int MAXN = 100010 ; //點數最大值    
const int MAXM = 400010 ; //邊數最大值    
const int INF = 0x3f3f3f3f;    
const long long int LLINF=1e16;

int S,V,F,P,ISUM;  
struct Edge{    
    int to,next,cap,flow;    
}edge[MAXM];//注意是MAXM    
int tol;    
int head[MAXN];    
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];    
long long int imap[555][555],IMAX;
int aa[555],bb[555]; 
void floyd(){  
    int i,j,k;  
    for(k=1;k<=F;k++){  
        for(i=1;i<=F;i++){  
            for(j=1;j<=F;j++){  
                imap[i][j]=imap[i][j]<(imap[i][k]+imap[k][j])?imap[i][j]:(imap[i][k]+imap[k][j]);
            }  
        }  
    } 
    //printf("%I64d",imap[1][20]);
/*
	printf("測試輸出\n"); 
    for(i=1;i<=F;i++){ 
        for(j=1;j<=F;j++){ 
            printf("%I64d  ",imap[i][j]); 
        } 
        printf("\n"); 
    }*/ 
}  
  
void init(){    
    tol = 0;    
    memset(head,-1,sizeof(head));    
}     
  
void addedge(int u,int v,int w,int rw=0){    
    edge[tol].to = v;    
    edge[tol].cap = w;    
    edge[tol].next = head[u];    
    edge[tol].flow = 0;    
    head[u] = tol++;    
    edge[tol].to = u;    
    edge[tol].cap = rw;    
    edge[tol].next = head[v];    
    edge[tol].flow = 0;    
    head[v] = tol++;    
}    
  //最大流開始   
int sap(int start,int end,int N){    
    memset(gap,0,sizeof(gap));    
    memset(dep,0,sizeof(dep));    
    memcpy(cur,head,sizeof(head));    
    int u = start;    
    pre[u] = -1;    
    gap[0] = N;    
    int ans = 0;    
    while(dep[start] < N){    
        if(u==end){    
            int Min = INF;    
            for(int i=pre[u];i!= -1; i=pre[edge[i^1].to])    
                if(Min > edge[i].cap - edge[i].flow)    
                    Min = edge[i].cap - edge[i].flow;    
                        
            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to]){    
                edge[i].flow += Min;    
                edge[i^1].flow -=Min;    
            }    
            u=start;    
            ans +=Min;    
            continue;    
        }    
        bool flag = false;    
        int v;    
        for(int i= cur[u];i!=-1;i=edge[i].next){    
            v=edge[i].to;    
            if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u]){    
                flag=true;    
                cur[u]=pre[v]=i;    
                break;    
            }    
        }    
        if(flag){    
            u=v;    
            continue;    
        }    
        int Min = N;    
        for(int i=head[u];i!= -1;i=edge[i].next)    
            if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min){    
                Min=dep[edge[i].to];    
                cur[u] = i;    
            }    
        gap[dep[u]]--;    
        if(!gap[dep[u]]) return ans;    
        dep[u] = Min +1;    
        gap[dep[u]]++;    
        if(u!=start) u = edge[pre[u]^1].to;    
    }    
    return ans;    
}    
//最大流結束   

bool build(long long int mid){
	int i,j,k,l,m,n;
	init();
	
	for(i=1;i<=F;i++){
		addedge(S,i,aa[i]);
		addedge(i+F,V,bb[i]);
	}
	for(i=1;i<=F;i++)
		for(j=1;j<=F;j++){
			if(imap[i][j]<=mid)	addedge(i,j+F,INF);
		}
	
	int orz=sap(S,V,V+1);
	return orz==ISUM;
	
}



long long int solve(){  
    int i,j,k;
	for(i=1;i<=F;i++){  
    	for(j=1;j<=F;j++){  
       		if(imap[i][j]!=LLINF){
        		IMAX=IMAX>imap[i][j]?IMAX:imap[i][j];
          	}
		}  
  	}  
    long long int l=0,r=IMAX,mid,ans=-1;
    
    while(r>=l){
    	mid=(l+r)>>1;
    	if(build(mid)){
	    	r=mid-1;
	    	ans=mid;
	    }
	    else{
    		l=mid+1;
    	}
    }
    return ans;
    
}  
  
int main(){    
    int m,n,q,p;    
    int i,j,k,a,b,c; 
	for(i=0;i<555;i++){
		for(j=0;j<555;j++){
			if(i==j)	imap[i][j]=0;
			else imap[i][j]=LLINF;
		}
	}
	IMAX=0;
    scanf("%d%d",&F,&P);
	S=0;
	V=2*F+1; 
	ISUM=0;
  	for(i=1;i<=F;i++){
	  	scanf("%d%d",&aa[i],&bb[i]);
	  	ISUM+=aa[i];
	}
	for(i=1;i<=P;i++){
		scanf("%d%d%d",&a,&b,&c);
		if(imap[a][b]>c){
			imap[a][b]=imap[b][a]=c;
		}
	}
	floyd();
	long long int orz=solve();
	printf("%lld\n",orz);
	return 0;
}