Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 16756 | Accepted: 3656 |
Description
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
* 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
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
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.
题意是有F个牛棚。里面现在有多少只牛 但是最多能容纳多少只牛 所以其他牛需要去别的棚子
棚子和棚子之间是无向图。。。用floyd一遍就搞定了
这道题。。不知道怎么回事。。无脑RE 重写到第三遍的时候。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 ; //点数最大值
const int MAXM = 400010 ; //边数最大值
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("测试输出\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++;
}
//最大流开始
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;
}
//最大流结束
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;
}