USACO 4.4 Pollutant Control

非常不错的一个最大流的题目，对边的处理很巧妙。这个链接的解题报告比较好：http://hi.baidu.com/mengyun1993/blog/item/c30d193c9a85932870cf6cda.html

总体上分3步，这题不看解题报告真不好做，我自己想了半个多小时，只想到了如何判断一条边是否在最小割中，也就是解题报告中说的“退流定理”～

再赞一下USACO，题目确实很不错，很多问题很值得花时间好好研究一下。

 

Test 1: TEST OK [0.000 secs, 3224 KB]
   Test 2: TEST OK [0.000 secs, 3224 KB]
   Test 3: TEST OK [0.000 secs, 3092 KB]
   Test 4: TEST OK [0.000 secs, 3224 KB]
   Test 5: TEST OK [0.000 secs, 3224 KB]
   Test 6: TEST OK [0.000 secs, 3224 KB]
   Test 7: TEST OK [0.000 secs, 3224 KB]
   Test 8: TEST OK [0.000 secs, 3224 KB]
   Test 9: TEST OK [0.000 secs, 3224 KB]
   Test 10: TEST OK [0.022 secs, 3224 KB]
   Test 11: TEST OK [0.000 secs, 3224 KB]
   Test 12: TEST OK [0.000 secs, 3224 KB]

/*
ID: zlqest11
LANG: C++
TASK: milk6
*/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

typedef long long llg;

const int N = 35;
const int M = 1010;
const llg INF = 1LL<<60;

struct node{
    int u, v, w;
}edge[M];

int n, m;
llg miniCut, c[N][N], f[N][N], tf[N][N];
//data structure for dinic
int lab[N], queue[N];

vector <int> ans;

bool bfs(int s, int t){
    int u, head, tail;
    memset(lab, -1, sizeof(lab));
    head = tail = lab[s] = 0;
    queue[0] = s;
    while(head <= tail){
        u = queue[head++];
        for(int i = 1; i <= n; i++){
            if(c[u][i]>f[u][i] && lab[i]==-1){
                lab[i] = lab[u] + 1;
                queue[++tail] = i;
            }
        }
        if(lab[t] != -1)  return  true;
    }
    return  false;
}

llg dinic_dfs(int u, llg delta){
    llg sum, tmp;
    if(u == n)  return  delta;
    sum = 0;
    for(int i = 1; i <= n; i++){
        if(lab[i]==lab[u]+1 && c[u][i]>f[u][i]){
            tmp = dinic_dfs(i, min(delta, c[u][i]-f[u][i]));
            delta -= tmp;
            sum += tmp;
            f[u][i] += tmp;
            f[i][u] -= tmp;
        }
    }
    return  sum;
}

llg maxFlow(){
    llg ans = 0;
    while(bfs(1, n)){
        ans += dinic_dfs(1, INF);
    }
    return  ans;
}

void makeTmp(){
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            tf[i][j] = f[i][j];
}

void reTmp(){
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            f[i][j] = tf[i][j];
}

int main()
{
    int a, b, w;
    freopen("milk6.in", "r", stdin);
    freopen("milk6.out", "w", stdout);
    //get data and init graph
    scanf("%d%d", &n, &m);
    memset(c, 0, sizeof(c));
    for(int i = 0; i < m; i++){
        scanf("%d%d%d", &a, &b, &w);
        edge[i].u = a;
        edge[i].v = b;
        edge[i].w = (llg)w*(m+1)+1;
        c[a][b] += edge[i].w;
    }
    memset(f, 0, sizeof(f));
    miniCut = maxFlow();
    ans.clear();
    llg xCut = miniCut, yCut;
    for(int i = 0; i < m; i++){
        a = edge[i].u;
        b = edge[i].v;
        w = edge[i].w;
        if(c[a][b] == f[a][b]){
            //tmp for flow
            makeTmp();
            c[a][b] -= w;
            memset(f, 0, sizeof(f));
            yCut = maxFlow();
            if(xCut-yCut == w){
                ans.push_back(i+1);
                xCut = yCut;
            }
            else{
                c[a][b] += w;
                reTmp();
            }
        }
    }
    printf("%d %d\n", (int)(miniCut/(m+1)), (int)(miniCut%(m+1)));
    for(int i = 0; i < ans.size(); i++)
        printf("%d\n", ans[i]);
    return 0;
}