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bzoj 1877 最小费用流

时间:2018-09-05 09:12:48      阅读:144      评论:0      收藏:0      [点我收藏+]

标签:bzoj   int   fir   bsp   from   rom   cond   pop   push   

思路:挺裸的费用流,拆拆点就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg
using namespace std;

const int N = 500 + 7;
const int M = 5e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1000000007;

int n, m, edgenum, S, T, head[N], dist[N], pre[N];
bool vis[N];

struct Edge {
    int from, to, cap, flow, cost, next;
} edge[M];

void init() {
    edgenum = 0;
    memset(head, -1, sizeof(head));
}

void addEdge(int u, int v, int w, int c) {
    Edge E1 = {u, v, w, 0, c, head[u]};
    edge[edgenum] = E1;
    head[u] = edgenum++;
    Edge E2 = {v, u, 0, 0, -c, head[v]};
    edge[edgenum] = E2;
    head[v] = edgenum++;
}

bool SPFA(int s, int t) {
    queue<int> Q;
    memset(dist, INF, sizeof(dist));
    memset(vis, false, sizeof(vis));
    memset(pre, -1, sizeof(pre));
    dist[s] = 0; vis[s] = true; Q.push(s);
    while(!Q.empty()) {
        int u = Q.front(); Q.pop(); vis[u] = false;
        for(int i = head[u]; i != -1; i = edge[i].next) {
            Edge E = edge[i];
            if(dist[E.to] > dist[u] + E.cost && E.cap > E.flow) {
                dist[E.to] = dist[u] + E.cost;
                pre[E.to] = i;
                if(!vis[E.to]) {
                    vis[E.to] = true;
                    Q.push(E.to);
                }
            }
        }
    }
    return pre[t] != -1;
}

void MCMF(int s, int t, int &cost, int &flow) {
    flow = 0; cost = 0;
    while(SPFA(s, t)) {
        int Min = INF;
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) {
            Edge E = edge[i];
            Min = min(Min, E.cap - E.flow);
        }
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
}
int main() {
    init();
    scanf("%d%d", &n, &m);
    S = 1, T = 2 * n;
    for(int i = 2; i < n; i++) addEdge(i, i + n, 1, 0);
    addEdge(1, 1 + n, inf, 0); addEdge(n, n + n, inf, 0);
    for(int i = 1; i <= m; i++) {
        int u, v, c;
        scanf("%d%d%d", &u, &v, &c);
        addEdge(u + n, v, 1, c);
    }

    int cost, flow;
    MCMF(S, T, cost, flow);
    printf("%d %d\n", flow, cost);
    return 0;
}

/*
*/

 

bzoj 1877 最小费用流

标签:bzoj   int   fir   bsp   from   rom   cond   pop   push   

原文地址:https://www.cnblogs.com/CJLHY/p/9589430.html

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