bzoj1061: [Noi2008]志愿者招募

orz

题解戳这里 https://www.byvoid.com/blog/noi-2008-employee/

暂时只能理解下

由于不等式不能作差，加上Yi变成等式作差

作差后每个X最多出现一次，这样就可以考虑两个之间的影响了（通过连边，这里流出去这么多那里就要流进来这么对，对应一正一负）

 

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<algorithm>
  5 #include<iostream>
  6 
  7 using namespace std;
  8 
  9 void setIO(const string& s) {
 10     freopen((s + ".in").c_str(), "r", stdin);
 11     freopen((s + ".out").c_str(), "w", stdout);
 12 }
 13 template<typename Q> Q read(Q& x) {
 14     static char c, f;
 15     for(f = 0; c = getchar(), !isdigit(c); ) if(c == '-') f = 1;
 16     for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - '0';
 17     if(f) x = -x;
 18     return x;
 19 }
 20 template<typename Q> Q read() {
 21     static Q x; read(x); return x;
 22 }
 23 // MCMF
 24 
 25 const int N = 1000 + 10, M = 25000 + 10, INF = ~0u >> 1;
 26 
 27 struct Edge {
 28     int to, adv, cost;
 29     Edge *next;
 30     Edge(int to = 0, int adv = 0, int cost = 0, Edge *next = 0) : to(to), adv(adv), cost(cost), next(next) {}
 31 }pool[M * 2], *pis = pool, *fir[N];
 32 
 33 void AddEdge(int u, int v, int w, int c) {
 34     fir[u] = new(pis++) Edge(v, w, c, fir[u]);
 35     fir[v] = new(pis++) Edge(u, 0, -c, fir[v]);
 36 }
 37 
 38 #define inv(p) (pool + ( ((p) - pool)^1 ))
 39 
 40 namespace MCMF {
 41     int n, s, t;
 42     int d[N], q[N], ql, qr, a[N];
 43     Edge *pre[N];
 44     bool inq[N];
 45     
 46     bool insert(int u, int dis, int aug) {
 47         if(d[u] > dis) {
 48             a[u] = aug;
 49             d[u] = dis;
 50             if(!inq[u]) {
 51                 q[qr++] = u;
 52                 if(qr == N) qr = 0;
 53                 inq[u] = 1;
 54             }
 55             return 1;
 56         }return 0;
 57     }
 58     
 59     bool SPFA(int& flow, int& cost) {
 60         for(int u = 1; u <= n; u++) d[u] = INF, inq[u] = 0;
 61         insert(s, 0, INF);
 62         while(ql ^ qr) {
 63             int u = q[ql++]; if(ql == N) ql = 0; inq[u] = 0;
 64             for(Edge *p = fir[u]; p; p = p->next) if(p->adv) {
 65                 if(insert(p->to, d[u] + p->cost, min(a[u], p->adv))) {
 66                     pre[p->to] = p;
 67                 }
 68             }
 69         }
 70         if(d[t] == INF) return 0;
 71         flow += a[t];
 72         cost += d[t] * a[t];
 73         for(int u = t; u != s; u = inv(pre[u])->to) {
 74             pre[u]->adv -= a[t];
 75             inv(pre[u])->adv += a[t];
 76         }
 77         return 1;
 78     }
 79     
 80     int main(int _n ,int _s, int _t) {
 81         n = _n, s = _s, t = _t;
 82         int flow = 0, cost = 0;
 83         while(SPFA(flow, cost));
 84         return cost;
 85     }
 86 }
 87 
 88 const int maxn = 1000 + 10, maxm = 10000 + 10;
 89 int demand[maxn];
 90 
 91 int main() {
 92 #ifdef DEBUG
 93     freopen("in.txt", "r", stdin);
 94     freopen("out.txt", "w", stdout);
 95 #endif
 96     
 97     int n = read<int>(), m = read<int>();
 98     int s = n + 2, t = s + 1;
 99     for(int i = 1; i <= n; i++) {
100         read(demand[i]);
101         AddEdge(i + 1, i, INF, 0);
102     }
103     for(int i = 1; i <= n + 1; i++) {
104         int dlt = demand[i] - demand[i - 1];
105         if(dlt >= 0) AddEdge(s, i, dlt, 0);
106         else AddEdge(i, t, -dlt, 0);
107     }
108     int l, r, c;
109     for(int i = 1; i <= m; i++) {
110         read(l), read(r), read(c);
111         AddEdge(l, r + 1, INF, c);
112     }
113     
114     printf("%d\n", MCMF::main(t, s, t));
115     
116     return 0;
117 }
View Code

 

vijos1891类似

posted @ 2015-12-15 20:54  Showson  阅读(246)  评论(1)  编辑  收藏  举报