不难写。调了1hQAQ
1.莫名其妙的错误。不要太多的东西放在主函数里面写，写成一个一个的函数比较清晰；
2.r=inf-1。如果=inf很明显是错的。
然后就因为这两个原因WA了1h。sading 。。。
哦对了。这个是floyed+二分+maxflow.有个 饶齐CSDN 介绍很多东西。。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,n) for(int i=1;i<=n;i++)
#define ll long long
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
int x=0;char c=getchar();bool f=true;
while(!isdigit(c)) {
if(c=='-') f=false;c=getchar();
}
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return f?x:-x;
}
const int nmax=505;
const int maxn=100005;
const ll inf=1e15;
const int Inf=0x7f7f7f7f;
struct edge{
int to,cap;edge *next,*rev;
};
edge edges[maxn],*pt,*head[nmax],*cur[nmax],*p[nmax];
int cnt[nmax],h[nmax],a[nmax],b[nmax],F,sum=0;
ll g[nmax][nmax];
void add(int u,int v,int d){
pt->to=v;pt->cap=d;pt->next=head[u];head[u]=pt++;
}
void adde(int u,int v,int d){
add(u,v,d);add(v,u,0);head[u]->rev=head[v];head[v]->rev=head[u];
}
int maxflow(int s,int t,int n){
clr(cnt,0);cnt[0]=n;clr(h,0);clr(cur,0);
int flow=0,a=Inf,x=s;edge *e;
while(h[s]<n){
for(e=cur[x];e;e=e->next) if(e->cap>0&&h[x]==h[e->to]+1) break;
if(e){
a=min(a,e->cap),p[e->to]=cur[x]=e,x=e->to;
if(x==t){
while(x!=s) p[x]->cap-=a,p[x]->rev->cap+=a,x=p[x]->rev->to;
flow+=a,a=Inf;
}
}else{
if(!--cnt[h[x]]) break;
h[x]=n;
for(e=head[x];e;e=e->next) if(e->cap>0&&h[x]>h[e->to]+1)
h[x]=h[e->to]+1,cur[x]=e;
cnt[h[x]]++;
if(x!=s) x=p[x]->rev->to;
}
}
return flow;
}
ll find(){
ll l=0,r=inf-1,m;
int s=0,t=F+F+1,n=t+1;ll ans=-1;
while(l<=r){
m=(l+r)>>1;pt=edges;clr(head,0);
rep(i,F) adde(s,i,a[i]),adde(i+F,t,b[i]);
rep(i,F) rep(j,F) if(g[i][j]<=m) adde(i,j+F,Inf);
/*for(int i=0;i<=n;i++){
for(edge *o=head[i];o;o=o->next) printf("%d %d
",o->to,o->cap);
printf("
");
}*/
int tmp;
if((tmp=maxflow(s,t,n))==sum) r=m-1,ans=m;
else l=m+1;
// printf("%lld %lld %lld:%d
",l,r,m,tmp);
}
return ans;
}
int main(){
F=read();int P=read(),u,v;ll d;
rep(i,F) a[i]=read(),b[i]=read(),sum+=a[i];
rep(i,F) rep(j,F) g[i][j]=inf;
rep(i,F) g[i][i]=0;
rep(i,P) u=read(),v=read(),d=read(),g[u][v]=g[v][u]=min(d,g[u][v]);
rep(k,F) rep(i,F) rep(j,F) if(g[i][k]!=inf&&g[k][j]!=inf)
g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
printf("%lld
",find());
return 0;
}
1738: [Usaco2005 mar]Ombrophobic Bovines 发抖的牛
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 367 Solved: 134
[Submit][Status][Discuss]
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
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120
Sample Output
1号田的7只牛中，2只牛直接进入1号田的雨棚，4只牛进入1号田的雨棚，1只进入3号田的雨棚，加入其他的由3号田来的牛们．所有的牛都能在110单位时间内到达要去的雨棚．