zoj 月赛

 

 E题

  1 /*Author :usedrose  */

  2 /*Created Time :2015/7/26 22:29:35*/

  3 /*File Name :2.cpp*/

  4 #include <cstdio>

  5 #include <iostream>

  6 #include <algorithm>

  7 #include <sstream>

  8 #include <cstdlib>

  9 #include <cstring>

 10 #include <climits>

 11 #include <vector>

 12 #include <string>

 13 #include <ctime>

 14 #include <cmath>

 15 #include <deque>

 16 #include <queue>

 17 #include <stack>

 18 #include <set>

 19 #include <map>

 20 #define INF 0x3f3f3f3f

 21 #define eps 1e-8

 22 #define pi acos(-1.0)

 23 #define OK cout << "ok" << endl;

 24 #define o(a) cout << #a << " = " << a << endl

 25 #define o1(a,b) cout << #a << " = " << a << "  " << #b << " = " << b << endl

 26 using namespace std;

 27 typedef long long LL;

 28 

 29 //最小费用最大流，求最大费用只需要取相反数，结果取相反数即可。

 30 //点的总数为 N，点的编号 0~N-1

 31 const int MAXN = 10000;

 32 const int MAXM = 100000;

 33 

 34 int n,m;  

 35   

 36 int A[MAXN],B[MAXN],X[MAXN],Y[MAXN]; 

 37 

 38 struct Edge

 39 {

 40     int to,next,cap,flow,cost;

 41 }edge[MAXM];

 42 int head[MAXN],tol;

 43 int pre[MAXN],dis[MAXN];

 44 bool vis[MAXN];

 45 int N;//节点总个数，节点编号从0~N-1

 46 

 47 void init(int n)

 48 {

 49     N = n;

 50     tol = 0;

 51     memset(head,-1,sizeof(head));

 52 }

 53 

 54 void addedge(int u,int v,int cap,int cost)

 55 {

 56     edge[tol].to = v;

 57     edge[tol].cap = cap;

 58     edge[tol].cost = cost;

 59     edge[tol].flow = 0;

 60     edge[tol].next = head[u];

 61     head[u] = tol++;

 62     edge[tol].to = u;

 63     edge[tol].cap = 0;

 64     edge[tol].cost = -cost;

 65     edge[tol].flow = 0;

 66     edge[tol].next = head[v];

 67     head[v] = tol++;

 68 }

 69 

 70 bool spfa(int s,int t)

 71 {

 72     queue<int>q;

 73     for(int i = 0;i < N;i++)

 74     {

 75         dis[i] = INF;

 76         vis[i] = false;

 77         pre[i] = -1;

 78     }

 79     dis[s] = 0;

 80     vis[s] = true;

 81     q.push(s);

 82     while(!q.empty())

 83     {

 84         int u = q.front();

 85         q.pop();

 86         vis[u] = false;

 87         for(int i = head[u]; i != -1;i = edge[i].next)

 88         {

 89             int v = edge[i].to;

 90             if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost )

 91             {

 92                 dis[v] = dis[u] + edge[i].cost;

 93                 pre[v] = i;

 94                 if(!vis[v])

 95                 {

 96                     vis[v] = true;

 97                     q.push(v);

 98                 }

 99             }

100         }

101     }

102     if(pre[t] == -1) return false;

103     else return true;

104 }

105 

106 //返回的是最大流， cost存的是最小费用

107 int minCostMaxflow(int s,int t,int &cost)

108 {

109     int flow = 0;

110     cost = 0;

111     while(spfa(s,t))

112     {

113         int Min = INF;

114         for(int i = pre[t];i != -1;i = pre[edge[i^1].to])

115         {

116             if(Min > edge[i].cap - edge[i].flow)

117             Min = edge[i].cap - edge[i].flow;

118         }

119         for(int i = pre[t];i != -1;i = pre[edge[i^1].to])

120         {

121             edge[i].flow += Min;

122             edge[i^1].flow -= Min;

123             cost += edge[i].cost * Min;

124         }

125         flow += Min;

126     }

127     return flow;

128 }

129 

130 

131 int main()

132 {

133     //freopen("data.in","r",stdin);

134     //freopen("data.out","w",stdout);

135     cin.tie(0);

136     ios::sync_with_stdio(false);

137     while (cin >> n >> m) {

138         int all = 0;

139         init(n+2);

140         int s = 0, t = n+1;

141         for (int i = 1;i <= n; ++ i) {

142             cin >> A[i] >> B[i];

143             if (A[i] > B[i])

144                 addedge(s, i, A[i]-B[i], 0);

145             else if (A[i] < B[i]) {

146                 addedge(i, t, B[i]-A[i], 0);

147                 all += (B[i] - A[i]);

148             }

149         }

150         int u, v;

151         for (int i = 0;i < m; ++ i) {

152             cin >> u >> v;

153            addedge(u, v, INF, 1);

154            addedge(v, u, INF, 1);       

155         }

156         int cost, ans;

157         ans = minCostMaxflow(s, t, cost);

158         if (ans < all) cout << -1 << endl;

159         else cout << cost << endl;

160     }

161     return 0;

162 }

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