链接：http://wikioi.com/problem/1035/
怎么说呢，只能说这个建图很有意思。因为只有m条道，然后能互相接在一起的连通，对每个点进行拆点，很有意思的一道裸费用留题。
代码：
1 #include <iostream> 2 #include <queue> 3 #include <iostream> 4 #include <cstdio> 5 #include <cstring> 6 #include <vector> 7 using namespace std; 8 const int maxn = 350; 9 const int inf = 10000000; 10 struct node 11 { 12 int u,v,cap,flow,cost,next; 13 }edges[100050]; 14 int head[maxn],cnt; 15 void init(int n) 16 { 17 int i; 18 for(i = 0;i <= n;i++) 19 head[i] = -1; 20 cnt = 0; 21 22 return ; 23 } 24 void addedge(int u,int v,int cap,int flow,int cost) 25 { 26 edges[cnt].u = u; 27 edges[cnt].v = v; 28 edges[cnt].cap = cap; 29 edges[cnt].flow = flow; 30 edges[cnt].cost = cost; 31 edges[cnt].next = head[u]; 32 head[u] = cnt; 33 cnt++; 34 edges[cnt].u = v; 35 edges[cnt].v = u; 36 edges[cnt].cap = 0; 37 edges[cnt].flow = flow; 38 edges[cnt].cost = -cost; 39 edges[cnt].next = head[v]; 40 head[v] = cnt; 41 cnt++; 42 } 43 int vis[maxn],a[maxn],pre[maxn],dis[maxn]; 44 int spfa(int s,int t,int n,int &flow,int &cost) 45 { 46 int i; 47 queue<int> q; 48 for(i = 0;i <= n ;i++) 49 dis[i] = -1,vis[i] = 0; 50 51 dis[s] = 0; 52 pre[s] = 0; 53 vis[s] = 1; 54 a[s] = inf; 55 56 int u,v; 57 q.push(s); 58 59 while(!q.empty()) 60 { 61 u = q.front(); 62 q.pop(); 63 vis[u] = 0; 64 65 for(i = head[u];i != -1;i = edges[i].next) 66 { 67 struct node & e = edges[i]; 68 69 v = e.v; 70 if(e.cap > e.flow &&dis[v] < dis[u]+e.cost) 71 { 72 dis[v] = dis[u]+e.cost; 73 a[v] = min(a[u],e.cap-e.flow); 74 pre[v] = i; 75 if(!vis[v]) 76 { 77 vis[v] = 1; 78 q.push(v); 79 } 80 } 81 } 82 } 83 84 if(dis[t] <= 0) 85 return 0; 86 flow+= a[t]; 87 cost += dis[t]*a[t]; 88 u = t; 89 while(u != s) 90 { 91 edges[pre[u]].flow += a[t]; 92 edges[pre[u]^1].flow -= a[t]; 93 u = edges[pre[u]].u; 94 } 95 return 1; 96 } 97 int MCMF(int s,int t,int n) 98 { 99 int flow = 0,cost = 0; 100 101 while(spfa(s,t,n,flow,cost)); 102 103 return cost; 104 } 105 int r[150],c[150],s[150]; 106 int main() 107 { 108 int n,m,u,v,i,j; 109 110 scanf("%d %d",&n,&m); 111 init(3*m); 112 for(i = 1;i <= m;i++) 113 cin>>r[i]>>c[i]>>s[i]; 114 115 addedge(0,2*m+1,n,0,0); 116 for(i = 1;i <= m;i++) 117 { 118 addedge(2*m+1,i,inf,0,0); 119 addedge(i,m+i,1,0,c[i]); 120 addedge(m+i,2*m+2,inf,0,0); 121 for(j = 1;j <= m;j++) 122 { 123 if(r[i]+s[i] < r[j]) 124 addedge(m+i,j,inf,0,0); 125 } 126 } 127 double ans; 128 ans = 1.0*MCMF(0,2*m+2,2*m+3); 129 130 printf("%.2f ",ans/100); 131 return 0; 132 }