【网络流】 TOJ 3864. SanXI

这题和HDOJ4307相似。。。本质上就是一个最小割的模型。。。

http://blog.csdn.net/weiguang_123/article/details/8077385

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 405
#define maxm 800005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL; 
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

struct Edge
{
	int v, c, next;
	Edge() {}
	Edge(int v, int c, int next) : v(v), c(c), next(next) {}
}E[maxm];

queue<int> q;
int H[maxn], cntE;
int dis[maxn];
int cur[maxn];
int cnt[maxn];
int pre[maxn];
int n, s, t, flow, nv;

void addedges(int u, int v, int c)
{
	E[cntE] = Edge(v, c, H[u]);
	H[u] = cntE++;
	E[cntE] = Edge(u, 0, H[v]);
	H[v] = cntE++;
}

void bfs()
{
	memset(cnt, 0, sizeof cnt);
	memset(dis, -1, sizeof dis);
	cnt[0]++, dis[t] = 0;
	q.push(t);
	while(!q.empty()) {
		int u = q.front();
		q.pop();
		for(int e = H[u]; ~e; e = E[e].next) {
			int v = E[e].v;
			if(dis[v] == -1) {
				dis[v] = dis[u] + 1;
				cnt[dis[v]]++;
				q.push(v);
			}
		}
	}
}

int isap()
{
	memcpy(cur, H, sizeof cur);
	bfs();
	flow = 0;
	int u = pre[s] = s, f, minv, e, pos;
	while(dis[s] < nv) {
		if(u == t) {
			f = INF;
			for(int i = s; i != t; i = E[cur[i]].v) if(f > E[cur[i]].c) {
				f = E[cur[i]].c;
				pos = i;
			}
			for(int i = s; i != t; i = E[cur[i]].v) {
				E[cur[i]].c -= f;
				E[cur[i] ^ 1].c += f;
			}
			flow += f;
			u = pos;
		}
		for(e = H[u]; ~e; e = E[e].next) if(E[e].c && dis[E[e].v] + 1 == dis[u]) break;
		if(~e) {
			cur[u] = e;
			pre[E[e].v] = u;
			u = E[e].v;
		}
		else {
			if(--cnt[dis[u]] == 0) break;
			for(minv = nv, e = H[u]; ~e; e = E[e].next) if(E[e].c && minv > dis[E[e].v]) {
				minv = dis[E[e].v];
				cur[u] = e;
			}
			dis[u] = minv + 1;
			cnt[dis[u]]++;
			u = pre[u];
		}
	}
	return flow;
}

void init()
{
	cntE = 0;
	memset(H, -1, sizeof H);
}

char ss[maxn];

void work()
{
	int tt, kk;
	scanf("%d", &n);
	s = 0, t = n + 1, nv = t + 1;
	for(int i = 1; i <= n; i++) {
		scanf("%s", ss);
		scanf("%d%d", &kk, &tt);
		if(kk == 0) addedges(i, t, tt);
		else addedges(s, i, tt);
	}
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			scanf("%d", &tt);
			if(i == j) continue;
			addedges(i, j, tt);
		}
	}
	printf("%d\n", isap());
}

int main()
{
	int _;
	scanf("%d", &_);
	while(_--) {
		init();
		work();
	}


	return 0;
}