bzoj 1449 费用流

思路:先把没有进行的场次规定双方都为负,对于x胜y负 变为x + 1胜 y - 1 负所需要的代价为 2 * C[ i ] * x  - 2 * D[ i ] * y + C[ i ] + D[ i ],

我们根据这个拆边建图,对于a和b进行的一场w, w流出的流量为1,并指向a 和 b,然后跑费用流。

#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 = 6000 + 7;
const int M = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

int n, m, win[N], lose[N], C[N], D[N], ans[N], cnt[N], S, T;
int head[N], pre[N], dist[N], edgenum;
bool vis[N];
PII a[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, LL &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 = 0, T = n + m + 1;
    for(int i = 1; i <= n; i++) {
        scanf("%d%d%d%d", &win[i], &lose[i], &C[i], &D[i]);
        cnt[i] = win[i] + lose[i];
    }

    for(int i = 1; i <= m; i++) {
        scanf("%d%d", &a[i].fi, &a[i].se);
        addEdge(S, i, 1, 0);
        addEdge(i, m + a[i].fi, 1, 0);
        addEdge(i, m + a[i].se, 1, 0);
        cnt[a[i].fi]++; cnt[a[i].se]++;
    }

    LL ans = 0;
    for(int i = 1; i <= n; i++) {
        ans += 1ll * C[i] * win[i] * win[i] + 1ll * D[i] * (cnt[i] - win[i]) * (cnt[i] - win[i]);
    }
    for(int i = 1; i <= n; i++) {
        int num = cnt[i] - lose[i] - win[i];
        int x = win[i], y = cnt[i] - win[i];

        while(num--) {
            addEdge(m + i, T, 1, 2 * C[i] * x - 2 * D[i] * y + C[i] + D[i]);
            x++; y--;
        }
    }

    LL cost; int flow;
    MCMF(S, T, cost, flow);
    printf("%lld\n", ans + cost);
    return 0;
}

/*
*/

 

posted @ 2018-08-10 17:13  NotNight  阅读(153)  评论(0编辑  收藏  举报