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51Nod-1442-士兵的旅行

2017-09-04 21:38 155 查看
ACM模版

描述



题解

典型的网络流问题,最大流,无向图,拆点建成有向图,判断是否满流即可。模版题……

代码

#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN = 1010;

struct edge
{
int to, cap, rev;

edge(int a, int b, int c)
{
to = a;
cap = b;
rev = c;
}
};

int level[MAXN];
int iter[MAXN];
vector<edge> veg[MAXN];

void add_edge(int from, int to, int cap)
{
veg[from].push_back(edge(to, cap, (int)veg[to].size()));
veg[to].push_back(edge(from, 0, (int)veg[from].size() - 1));
}

void bfs(int s)
{
memset(level, -1, sizeof(level));

queue<int> que;
level[s] = 0;
que.push(s);

while (!que.empty())
{
int v = que.front();
que.pop();
for (int i = 0; i < veg[v].size(); i++)
{
edge &e = veg[v][i];
if (e.cap > 0 && level[e.to] < 0)
{
level[e.to] = level[v] + 1;
que.push(e.to);
}
}
}
}

int dfs(int v, int t, int f)
{
if (v == t)
{
return f;
}

for (int &i = iter[v]; i < veg[v].size(); i++)
{
edge &e = veg[v][i];
if (e.cap > 0 && level[v] < level[e.to])
{
int d = dfs(e.to, t, min(f, e.cap));
if (d > 0)
{
e.cap -= d;
veg[e.to][e.rev].cap += d;
return d;
}
}
}

return 0;
}

int max_flow(int s, int t)
{
int flow = 0, f;
while (1)
{
bfs(s);
if (level[t] < 0)
{
return flow;
}

memset(iter, 0, sizeof(iter));
while ((f = dfs(s, t, INF)) > 0)
{
flow += f;
}
}
}

int n, m;
int A[MAXN];
int B[MAXN];

int main()
{
for (int i = 0; i < MAXN; i++)
{
veg[i].clear();
}

scanf("%d%d", &n, &m);

int sum1 = 0, sum2 = 0, a, b;
for (int i = 1; i <= n; i++)
{
scanf("%d", &a);
add_edge(i, i + n, INF);
add_edge(0, i, a);
sum1 += a;
}
for (int i = 1; i <= n; i++)
{
scanf("%d", &a);
add_edge(i + n, 2 * n + 1, a);
sum2 += a;
}

for (int i = 0; i < m; i++)
{
scanf("%d%d", &a, &b);
add_edge(a, b + n, INF);
add_edge(b, a + n, INF);
}

int ans = max_flow(0, 2 * n + 1);
if (ans == sum2 && sum1 == sum2)
{
printf("YES\n");
}
else
{
printf("NO\n");
}

return 0;
}
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标签:  网络流 最大流