UVA 10330 Power Transmission
2024-09-05 22:07:19
题意:懒得打了。LUCKY CAT 里有 http://163.32.78.26/homework/q10330.htm
第一个网络流题目。每个节点都有一个容量值。需要拆点。拆成i - > i + N 边权为容量值
另外注意B个点 连接方式:s - 集合B
D个点 链接方式 集合D + N -> t汇点
其他看处理就好
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == ? b : gcd(b, a % b);}
#define MAXN 250
const int INF = 0x3f3f3f3f ;
int flow[MAXN][MAXN],c[MAXN][MAXN];
int p[MAXN];
int N,M,B,D,src,tag;
void read()
{
memset(c,,sizeof(c));
src = ; tag = * N + ;
for (int i = ; i <= N; i++)
{
int tmp;
scanf("%d",&tmp);
c[i][i + N] = tmp;
}
scanf("%d",&M);
for (int i = ; i <= M; i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
c[u + N][v] = w;
}
scanf("%d%d",&B,&D);
for (int i = ; i <= B; i++)
{
int tmp;
scanf("%d",&tmp);
c[][tmp] = INF;
}
for (int i = ; i <= D; i++)
{
int tmp;
scanf("%d",&tmp);
c[tmp + N][tag] = INF;
}
}
int Edmonds_karp(int src,int tag)
{
memset(flow,,sizeof(flow));
int ans = ;
queue<int>q;
while (!q.empty()) q.pop();
int a[MAXN];
while (true)
{
memset(a,,sizeof(a));
a[src] = INF;
q.push(src);
while (!q.empty())
{
int u = q.front(); q.pop();
for (int v = ; v <= tag; v++)
if (!a[v] && c[u][v] > flow[u][v])
{
p[v] = u;
q.push(v);
a[v] = min(a[u],c[u][v] - flow[u][v]);
}
}
if (a[tag] == ) break;
for (int u = tag; u != src; u = p[u])
{
flow[p[u]][u] += a[tag];
flow[u][p[u]] -= a[tag];
}
ans += a[tag];
}
return ans;
}
int main()
{
//freopen("sample.txt","r",stdin);
while (scanf("%d",&N)!=EOF)
{
read();
printf("%d\n",Edmonds_karp(src,tag));
}
return ;
}
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