uva 12264 Risk

https://vjudge.net/problem/UVA-12264

题意：

有很多个阵地，分为敌方和己方，每个士兵可以移动到相邻的己方的阵地，但是只能移动一步。
现在要让与敌方相邻的阵地中士兵最少的最多。

思路：

最少的最多，那显然二分。
二分这个最少的值。拆点，敌方阵地不管，S向己方阵地(i)向连边，容量为本阵地士兵的数量，(i')向T连边，如果是与敌方相邻的阵地，那么容量为二分的值；如果是处于己方阵地的包围，那么容量为1即可。最后跑最大流判断是否满流。

STD：

本STD在uva上AC，uvalive上WA，请谨慎食用。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 205;
int a[N];
char s[N][N];
bool is[N];

struct edge
{
	int u,v,cap;
	edge(int u,int v,int cap):u(u),v(v),cap(cap){}
	edge(){}
};

vector<edge> es;
vector<int> G[N];
int n,S,T;
int dis[N],cur[N];

void adde(int u,int v,int cap)
{
	es.push_back(edge(u,v,cap));
	es.push_back(edge(v,u,0));
	int sz = es.size();
	G[u].push_back(sz-2);
	G[v].push_back(sz-1);
}

bool bfs()
{
	memset(dis,inf,sizeof dis);
	dis[S] = 0;
	queue<int> q;
	q.push(S);
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		for (int i = 0;i < G[u].size();i++)
		{
			edge &e = es[G[u][i]];
			int v = e.v;
			if (e.cap > 0 && dis[v] >= inf)
			{
				dis[v] = dis[u] + 1;
				q.push(v);
			}
		}
	}
	return dis[T] < inf;
}

int dfs(int u,int flow)
{
	if (u == T) return flow;
	for (int i = cur[u];i < G[u].size();i++)
	{
		cur[u] = i;
		edge &e = es[G[u][i]];
		int v = e.v;
		if (dis[v] == dis[u] + 1 && e.cap > 0)
		{
			int tmp = dfs(v,min(e.cap,flow));
			if (tmp)
			{
				e.cap -= tmp;
				es[G[u][i]^1].cap += tmp;
				return tmp;
			}
		}
	}
	return 0;
}

int dinic()
{
	int ans = 0;
	while (bfs())
	{
		int tmp;
		memset(cur,0,sizeof(cur));
		while (tmp = dfs(S,inf)) ans += tmp;
	}
	return ans;
}

bool meet(int lim)
{
	for (int i = 0;i < N;i++) G[i].clear();
	es.clear();
	for (int i = 1;i <= n;i++)
	{
		if (a[i] <= 0) continue;
		adde(S,i,a[i]);
		adde(i,i+n,inf);
	}
	int sum = 0;
	for (int i = 1;i <= n;i++)
	{
		if (a[i] <= 0) continue;
		if (is[i])
		{
			sum += lim;
			adde(i+n,T,lim);
		}
		else
		{
			sum++;
			adde(i+n,T,1);
		}
	}
	for (int i = 1;i <= n;i++)
	{
		for (int j = 1;j <= n;j++)
		{
			if (i == j) continue;
			if (a[i] <= 0 || a[j] <= 0) continue;
			if (s[i][j] == 'Y')
			{
				adde(i,j+n,inf);
			}
		}
	}
	int ans = dinic();
	return ans >= sum;
}

int main()
{
	int t;
	scanf("%d",&t);
	while (t--)
	{
		memset(is,0,sizeof is);
		scanf("%d",&n);
		S = 0;
		T = 2 * n + 1;
		for (int i = 1;i <= n;i++)
		{
			scanf("%d",&a[i]);
		}
		for (int i = 1;i <= n;i++)
		{
			scanf("%s",s[i]+1);
		}
		for (int i = 1;i <= n;i++)
		{
			for (int j = 1;j <= n;j++)
			{
				if (a[i] > 0 && a[j] <= 0)
				{
					if (s[i][j] == 'Y') is[i] = true;
				}
			}
		}
		int l = 1,r = 1e5;
		while (r - l > 1)
		{
			int mid = (l + r) >> 1;
			if (meet(mid)) l = mid;
			else r = mid;
		}
		while (meet(l+1)) l++;
		printf("%d
",l);
	}
	return 0;
}

/*
3
1 1 0
NYN
YNY
NYN
7 
7 3 3 2 0 0 5 
NYNNNNN 
YNYYNNN 
NYNYYNN 
NYYNYNN 
NNYYNNN 
NNNNNNY 
NNNNNYN
4
2 2 0 0
NNYN
NNNY
YNNN
NYNN
*/