题意:两遍最长路,不能走重复点。和UVA 10806类似。
分析:拆点,u->u',MCMF,求的是最大流的最小费用,那么cost取负。
注意的是源点,源点不用拆,那么走出来的最小费用,左上角的点,右下角的点走了两遍,输出除去即可。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int maxn = 2500+5; 6 const int INF = 0x3f3f3f3f; 7 8 struct Edge 9 { 10 int from, to, cap, flow, cost; 11 }; 12 13 struct MCMF 14 { 15 int n, m; 16 vector<Edge> edges; 17 vector<int> G[maxn]; 18 bool inq[maxn]; // 是否在队列中 19 int d[maxn]; // Bellman-Ford 20 int p[maxn]; // 上一条弧 21 int a[maxn]; // 可改进量 22 23 void init(int n) 24 { 25 this->n = n; 26 for(int i = 0; i < n; i++) G[i].clear(); 27 edges.clear(); 28 } 29 30 void AddEdge(int from, int to, int cap, int cost) 31 { 32 edges.push_back((Edge) 33 { 34 from, to, cap, 0, cost 35 }); 36 edges.push_back((Edge) 37 { 38 to, from, 0, 0, -cost 39 }); 40 m = edges.size(); 41 G[from].push_back(m-2); 42 G[to].push_back(m-1); 43 } 44 45 bool BellmanFord(int s, int t, int &flow, long long& cost) 46 { 47 memset(inq,0,sizeof(inq)); 48 for(int i=0;i<n;i++) 49 d[i] = INF; 50 d[s] = 0; 51 inq[s] = true; 52 p[s] = 0; 53 a[s] = INF; 54 55 queue<int> Q; 56 Q.push(s); 57 while(!Q.empty()) 58 { 59 int u = Q.front(); 60 Q.pop(); 61 inq[u] = false; 62 for(int i = 0; i < G[u].size(); i++) 63 { 64 Edge& e = edges[G[u][i]]; 65 if(e.cap > e.flow && d[e.to] > d[u] + e.cost) 66 { 67 d[e.to] = d[u] + e.cost; 68 p[e.to] = G[u][i]; 69 a[e.to] = min(a[u], e.cap - e.flow); 70 if(!inq[e.to]) 71 { 72 Q.push(e.to); 73 inq[e.to] = true; 74 } 75 } 76 } 77 } 78 if(d[t] == INF) return false; //s-t 不连通,失败退出 79 flow += a[t]; 80 cost += (long long)d[t] * (long long)a[t]; 81 int u = t; 82 while(u != s) 83 { 84 edges[p[u]].flow += a[t]; 85 edges[p[u]^1].flow -= a[t]; 86 u = edges[p[u]].from; 87 } 88 return true; 89 } 90 91 long long Mincost(int s, int t) 92 { 93 long long cost = 0; 94 int flow = 0; 95 while(BellmanFord(s, t, flow, cost)) { 96 if(flow==2) 97 break; 98 }; 99 return cost; 100 } 101 }sol; 102 103 int maps[maxn][maxn]; 104 int aa[maxn*maxn]; 105 106 int main() 107 { 108 //freopen("in.txt","r",stdin); 109 int n; 110 while(scanf("%d",&n)!=EOF) { 111 112 for(int i=0;i<n;i++) 113 for(int j=0;j<n;j++) 114 scanf("%d",&maps[i][j]); 115 116 int s = 0,t = n*n-1; 117 sol.init(n*n*2); 118 119 for(int i=0;i<n;i++) { 120 for(int j=0;j<n;j++) { 121 int id = i*n + j; 122 if(id!=s&&id!=t) 123 sol.AddEdge(id,id+n*n,1,-maps[i][j]); 124 if(id==s) { 125 sol.AddEdge(id,id+1,1,0); 126 sol.AddEdge(id,id+n,1,0); 127 } 128 else { 129 if(i<n-1) sol.AddEdge(id+n*n,id+n,1,0); 130 if(j<n-1) sol.AddEdge(id+n*n,id+1,1,0); 131 } 132 } 133 } 134 135 printf("%d ",-sol.Mincost(s,t)+maps[0][0]+maps[n-1][n-1]); 136 137 138 139 } 140 return 0; 141 }