HDU 3435 A new Graph Game （KM）

题意:

N<=1000,M<=10000,删边求汉密顿圈的最小权

分析:

圈嘛,每个点只能属于一个圈,然后可以想到匹配了

怎么做呢,圈每个点有唯一的后继,然后拆点咯

判断可行看是不是完美匹配就好了,图大用km咯

代码:

//
//  Created by TaoSama on 2015-10-31
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1000 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, w

;
int match
, lx
, ly
, slack
;
bool S
, T
;

bool dfs(int i) {
S[i] = true;
for(int j = 1; j <= n; ++j) {
if(T[j]) continue;
if(lx[i] + ly[j] == w[i][j]) {
T[j] = true;
if(!match[j] || dfs(match[j])) {
match[j] = i;
return true;
}
} else slack[j] = min(slack[j], lx[i] + ly[j] - w[i][j]);
}
return false;
}

bool update() {
int a = INF;
for(int j = 1; j <= n; ++j)
if(!T[j]) a = min(a, slack[j]);
for(int i = 1; i <= n; ++i) {
if(S[i]) lx[i] -= a;
if(T[i]) ly[i] += a;
else slack[i] -= a;
}
return a != INF;
}

int km() {
for(int i = 1; i <= n; ++i) {
match[i] = lx[i] = ly[i] = 0;
for(int j = 1; j <= n; ++j)
lx[i] = max(lx[i], w[i][j]);
}
for(int i = 1; i <= n; ++i) {
memset(slack, 0x3f, sizeof slack);
while(true) {
memset(S, false, sizeof S);
memset(T, false, sizeof T);
if(dfs(i)) break;
else if(!update()) return 0;
}
}
int ret = 0;
for(int i = 1; i <= n; ++i) if(match[i]) ret += w[match[i]][i];
return ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
int kase = 0;
while(t--) {
scanf("%d%d", &n, &m);
memset(w, 0xc0, sizeof w);
for(int i = 1; i <= m; ++i) {
int u, v, c; scanf("%d%d%d", &u, &v, &c);
//u, u + n, v, v + n
w[v][u] = w[u][v] = max(w[u][v], -c);
}
int ans = -km();
if(ans) printf("Case %d: %d\n", ++kase, ans);
else printf("Case %d: NO\n", ++kase);
}
return 0;
}