vjudge传送门[here]
题目大意:给一个有(3≤v≤1000)个点e（3≤e≤10000）条边的有向加权图，求1~v的两条不相交（除了起点和终点外没有公共点）的路径，使权值和最小。
正解是吧2到v-1的每个点拆成两个点，中间连一条容量为1，费用为0的边，然后求1到v的流量为2的最小费用流就行了。
Code
1 /** 2 * Uva 3 * Problem#1658 4 * Accepted 5 */ 6 #include<iostream> 7 #include<cstdio> 8 #include<cctype> 9 #include<cstring> 10 #include<cstdlib> 11 #include<fstream> 12 #include<sstream> 13 #include<algorithm> 14 #include<map> 15 #include<set> 16 #include<queue> 17 #include<vector> 18 #include<stack> 19 using namespace std; 20 typedef bool boolean; 21 #define INF 0xfffffff 22 #define smin(a, b) a = min(a, b) 23 #define smax(a, b) a = max(a, b) 24 template<typename T> 25 inline boolean readInteger(T& u){ 26 char x; 27 int aFlag = 1; 28 while(!isdigit((x = getchar())) && x != '-' && ~x); 29 if(x == -1){ 30 return false; 31 } 32 if(x == '-'){ 33 x = getchar(); 34 aFlag = -1; 35 } 36 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 37 ungetc(x, stdin); 38 u *= aFlag; 39 return true; 40 } 41 42 ///map template starts 43 typedef class Edge{ 44 public: 45 int end; 46 int next; 47 int cap; 48 int flow; 49 int cost; 50 Edge(const int end = 0, const int next = 0, const int cap = 0, const int flow = 0, const int cost = 0):end(end), next(next), cap(cap), flow(flow), cost(cost){} 51 }Edge; 52 53 typedef class MapManager{ 54 public: 55 int ce; 56 int *h; 57 Edge *edge; 58 MapManager(){} 59 MapManager(int points, int limit):ce(0){ 60 h = new int[(const int)(points + 1)]; 61 edge = new Edge[(const int)(limit + 1)]; 62 memset(h, 0, sizeof(int) * (points + 1)); 63 } 64 inline void addEdge(int from, int end, int cap, int flow, int cost){ 65 edge[++ce] = Edge(end, h[from], cap, flow, cost); 66 h[from] = ce; 67 } 68 inline void addDoubleEdge(int from, int end, int cap, int cost){ 69 addEdge(from, end, cap, 0, cost); 70 addEdge(end, from, cap, cap, -cost); 71 } 72 Edge& operator [](int pos){ 73 return edge[pos]; 74 } 75 inline int reverse(int pos){ 76 return (pos & 1) ? (pos + 1) : (pos - 1); 77 } 78 inline void clean(){ 79 delete[] h; 80 delete[] edge; 81 ce = 0; 82 } 83 }MapManager; 84 #define m_begin(g, i) (g).h[(i)] 85 /// map template ends 86 87 int n, m; 88 MapManager g; 89 90 inline boolean init(){ 91 if(!readInteger(n)) return false; 92 readInteger(m); 93 g = MapManager(n * 2, m * 2 + n * 2); 94 for(int i = 1, a, b, w; i <= m; i++){ 95 readInteger(a); 96 readInteger(b); 97 readInteger(w); 98 g.addDoubleEdge(a + n, b, 1, w); 99 } 100 for(int i = 1; i <= n; i++){ //拆点 101 g.addDoubleEdge(i, i + n, 1, 0); 102 } 103 return true; 104 } 105 106 int* dis; 107 boolean* visited; 108 int* last; 109 int* laste; 110 int* mflow; 111 int cost; 112 int s, t; 113 queue<int> que; 114 int sizee; 115 116 inline void spfa(){ 117 memset(dis, 0x7f, sizeof(int) * (sizee)); 118 memset(visited, false, sizeof(boolean) * (sizee)); 119 que.push(s); 120 last[s] = 0; 121 dis[s] = 0; 122 laste[s] = 0; 123 mflow[s] = INF; 124 visited[s] = true; 125 while(!que.empty()){ 126 int e = que.front(); 127 que.pop(); 128 visited[e] = false; 129 for(int i = m_begin(g, e); i != 0; i = g[i].next){ 130 int &eu = g[i].end; 131 if(dis[e] + g[i].cost < dis[eu] && g[i].flow < g[i].cap){ 132 dis[eu] = dis[e] + g[i].cost; 133 last[eu] = e; 134 laste[eu] = i; 135 mflow[eu] = min(g[i].cap - g[i].flow, mflow[e]); 136 if(!visited[eu] && eu != t){ 137 que.push(eu); 138 visited[eu] = true; 139 } 140 } 141 } 142 } 143 for(int i = t; i != s; i = last[i]){ 144 g[laste[i]].flow += mflow[t]; 145 g[g.reverse(laste[i])].flow -= mflow[t]; 146 cost += mflow[t] * g[laste[i]].cost; 147 } 148 } 149 150 inline void minCostFlow(){ 151 s = 1 + n, t = n, sizee = 2 * n + 1; 152 dis = new int[(const int)(sizee)]; 153 visited = new boolean[(const int)(sizee)]; 154 last = new int[(const int)(sizee)]; 155 laste = new int[(const int)(sizee)]; 156 mflow = new int[(const int)(sizee)]; 157 spfa(); 158 spfa(); 159 } 160 161 inline void solve(){ 162 cost = 0; 163 minCostFlow(); 164 printf("%d ", cost); 165 g.clean(); 166 } 167 168 int main(){ 169 while(init()){ 170 solve(); 171 } 172 return 0; 173 }