1.题面
http://acm.hdu.edu.cn/showproblem.php?pid=4280
2.题意
给你网络,让你求最大流,数据有点大,需要比较靠谱的模板
3.思路
裸题,但是把我自己写的dinic干掉了
4.代码
/*
最大流模板
dinic算法
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=100 + 100000;//点数的最大值
const int MAXM=100 + 200000;//边数的最大值
struct Node
{
int from,to,next;
int cap;
}edge[MAXM];
int tol;
int dep[MAXN];//dep为点的层次
int head[MAXN];
int n;
void init()
{
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)//第一条变下标必须为偶数
{
edge[tol].from=u;
edge[tol].to=v;
edge[tol].cap=w;
edge[tol].next=head[u];
head[u]=tol++;
edge[tol].from=v;
edge[tol].to=u;
edge[tol].cap=w;
edge[tol].next=head[v];
head[v]=tol++;
}
int BFS(int start,int end)
{
int que[MAXN];
int front,rear;
front=rear=0;
memset(dep,-1,sizeof(dep));
que[rear++]=start;
dep[start]=0;
while(front!=rear)
{
int u=que[front++];
if(front==MAXN)front=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(edge[i].cap>0&&dep[v]==-1)
{
dep[v]=dep[u]+1;
que[rear++]=v;
if(rear>=MAXN)rear=0;
if(v==end)return 1;
}
}
}
return 0;
}
int dinic(int start,int end)
{
int res=0;
int top;
int stack[MAXN];//stack为栈,存储当前增广路
int cur[MAXN];//存储当前点的后继
while(BFS(start,end))
{
memcpy(cur,head,sizeof(head));
int u=start;
top=0;
while(1)
{
if(u==end)
{
int min=INF;
int loc;
for(int i=0;i<top;i++)
if(min>edge[stack[i]].cap)
{
min=edge[stack[i]].cap;
loc=i;
}
for(int i=0;i<top;i++)
{
edge[stack[i]].cap-=min;
edge[stack[i]^1].cap+=min;
}
res+=min;
top=loc;
u=edge[stack[top]].from;
}
for(int i=cur[u];i!=-1;cur[u]=i=edge[i].next)
if(edge[i].cap!=0&&dep[u]+1==dep[edge[i].to])
break;
if(cur[u]!=-1)
{
stack[top++]=cur[u];
u=edge[cur[u]].to;
}
else
{
if(top==0)break;
dep[u]=-1;
u=edge[stack[--top]].from;
}
}
}
return res;
}
typedef pair<int, int> pir;
int main()
{
/*std::ios::sync_with_stdio(false);cin.tie(0);*/
int T;
scanf("%d", &T);
while (T--){
init();
int n, m;
scanf("%d%d", &n, &m);
pir ss(INF, 0), tt(-INF, n), tmp;
//# cout << "n m = " << n << " " << m<< endl;
for (int i = 1; i <= n; i++){
scanf("%d%d", &tmp.first, &tmp.second);
tmp.second = i;
ss = min(tmp, ss);
tt = max(tmp, tt);
}
int s = ss.second, t = tt.second;
while (m--){
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
addedge(a, b, c);
//# addedge(b, a, c);
}
//# cout << "s t " << s << " " << t << endl;
int ans = dinic(s, t);
printf("%d\n", ans);
}/**/
return 0;
}