HDU 4289 Control

Posted 2020-11-01 16-chq

对城市拆点 城市之间的容量就是 在这个城市设立关卡需要的费用 求最小割 跑一遍最大流

#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

const int inf = 0x3f3f3f3f;
const int maxn = 100005;

int n, m;//点数、边数
int sp, tp;//原点、汇点

struct node {
    int u;
    int v, next;
    int cap;
}mp[maxn];

int pre[maxn], dis[maxn], cur[maxn];//cur为当前弧优化，dis存储分层图中每个点的层数（即到原点的最短距离），pre建邻接表
int cnt = 0;

void init() {  //不要忘记初始化
    cnt = 0;
    memset(pre, -1, sizeof(pre));
}

void add(int u, int v, int w) { //加边
    mp[cnt].u = u;
    mp[cnt].v = v;
    mp[cnt].cap = w;
    mp[cnt].next = pre[u];
    pre[u] = cnt++;

    mp[cnt].u = v;
    mp[cnt].v = u;
    mp[cnt].cap = 0;
    mp[cnt].next = pre[v];
    pre[v] = cnt++;
}

bool bfs() {  //建分层图
    memset(dis, -1, sizeof(dis));
    queue<int>q;
    while(!q.empty())
        q.pop();
    q.push(sp);
    dis[sp] = 0;
    int u, v;
    while(!q.empty()) {
        u = q.front();
        q.pop();
        for(int i = pre[u]; i != -1; i = mp[i].next) {
            v = mp[i].v;
            if(dis[v] == -1 && mp[i].cap > 0) {
                dis[v] = dis[u] + 1;
                q.push(v);
                if(v == tp)
                    break;
            }
        }
    }
    return dis[tp] != -1;
}

int dfs(int u, int cap) {//寻找增广路
    if(u == tp || cap == 0)
    return cap;
    int res = 0, f;
    for(int i = cur[u]; i != -1; i = mp[i].next) {//
        int v = mp[i].v;
        if(dis[v] == dis[u] + 1 && (f = dfs(v, min(cap - res, mp[i].cap))) > 0) {
            mp[i].cap -= f;
            mp[i ^ 1].cap += f;
            res += f;
            if(res == cap)
                return cap;
        }
    }
    if(!res)
        dis[u] = -1;
    return res;
}

int dinic() {
    int ans = 0;
    while(bfs()) {
        for(int i = sp; i <= tp; i++)
            cur[i] = pre[i];
        ans += dfs(sp, inf);
    }
    return ans;
}



int main()
{
    int n, m;
    int t1, t2;
    while(scanf("%d%d", &n, &m)!=EOF) {
    init();
    sp = 0;
    tp = 2*n+2;
    scanf("%d%d", &t1, &t2);
    add(sp, t1, inf);
    add(n+t2, tp, inf);
    for(int i = 1; i <= n; i++) {
        int t; scanf("%d", &t);
        add(i, i+n, t);
    }
    for(int i = 0; i < m; i++) {
        int t1, t2;
        scanf("%d%d", &t1, &t2);
        add(n+t1, t2, inf);
        add(n+t2, t1, inf);
    }
    int f = dinic();
    printf("%d\n", f);
    }
    return 0;
}