题目
分析
相当于一个模板题
最小费用最大流，点权是费用，流量为1限制只能一个人走
考虑拆点，拆为入点与出点，建一个源点连起点，建一个汇点连终点
流量等于N就输出费用，小于N输出-1
题解：
代码
1 /*************************** 2 User:Mandy.H.Y 3 Language:c++ 4 Problem:maze 5 Algorithm: 6 Score: 7 ***************************/ 8 9 //做网络流记得考虑拆点啊 10 11 #include<bits/stdc++.h> 12 13 using namespace std; 14 15 const int maxn = 150; 16 const int maxint = 2100000000; 17 18 int n,m,K,size=1,cur[4100],val[410],N,tot,s,t,cost,maxflow; 19 //size从2开始 20 int first[410]; 21 bool vis[410]; 22 int dis[410]; 23 24 struct Edge{ 25 int v,nt,w,f; 26 }edge[4100]; 27 28 char *TT,*mo,but[(1 << 15) + 2]; 29 #define getchar() ((TT == mo && (mo = ((TT = but) + fread(but,1,1 << 15,stdin)),TT == mo)) ? -1 : *TT++) 30 template<class T>inline void read(T &x){ 31 x = 0;bool flag = 0;char ch = getchar(); 32 while(!isdigit(ch)) flag |= ch == '-',ch = getchar(); 33 while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48),ch = getchar(); 34 if(flag) x = -x; 35 } 36 37 template<class T>void putch(const T x){ 38 if(x > 9) putch(x / 10); 39 putchar(x % 10 | 48); 40 } 41 42 template<class T>void put(const T x){ 43 if(x < 0) putchar('-'),putch(-x); 44 else putch(x); 45 } 46 47 void file(){ 48 freopen("maze.in","r",stdin); 49 // freopen("maze.out","w",stdout); 50 } 51 52 void eadd(int u,int v,int w,int f){ 53 edge[++ size].v = v;edge[size].w = w;edge[size].f = f;edge[size].nt = first[u];first[u] = size; 54 edge[++ size].v = u;edge[size].w = -w;edge[size].f = 0;edge[size].nt = first[v];first[v] = size; 55 } 56 57 void readdata(){ 58 read(n);read(m); 59 tot = n*m; 60 for(int i = 0;i < n; ++ i) 61 for(int j = 0;j < m; ++ j){ 62 int id = i * m + j; 63 read(val[id]); 64 eadd(id,id + tot,val[id],1); 65 } 66 67 read(K); 68 for(int i = 1;i <= K; ++ i){ 69 int u,v,x,y; 70 read(u);read(v);read(x);read(y); 71 eadd((u-1)*m+(v-1) + tot,(x-1)*m+(y-1),0,1); 72 eadd((x-1)*m+(y-1) + tot,(u-1)*m+(v-1),0,1); 73 } 74 75 read(N); 76 s = tot + tot + 1; 77 t = tot + tot + 2; 78 for(int i = 1;i <= N; ++ i){ 79 int x,y; 80 read(x);read(y); 81 eadd(s,(x-1)*m+(y-1),0,1); 82 } 83 for(int i = 1;i <= N; ++ i){ 84 int x,y; 85 read(x);read(y); 86 eadd((x-1)*m+(y-1)+tot,t,0,1); 87 } 88 } 89 90 bool SPFA(){ 91 memset(dis,0x3f3f3f3f,sizeof(dis)); 92 memset(vis,0,sizeof(vis)); 93 queue<int>q; 94 dis[s] = 0; 95 q.push(s); 96 while(!q.empty()){ 97 int u = q.front(); 98 q.pop(); 99 vis[u] = 0; 100 for(int i = first[u];i;i = edge[i].nt){ 101 int v = edge[i].v; 102 int w = edge[i].w; 103 if(dis[v] > dis[u] + w && edge[i].f){ 104 dis[v] = dis[u] + w; 105 if(!vis[v]) q.push(v),vis[v] = 1; 106 } 107 } 108 } 109 return dis[t] != 0x3f3f3f3f; 110 } 111 112 int dfs(int u,int flow){ 113 if(u == t || flow == 0) return flow; 114 vis[u] = 1; 115 int left = flow; 116 117 for(int i = first[u];i;i = edge[i].nt){ 118 int v = edge[i].v; 119 if(edge[i].f && (!vis[v]) && dis[v] == dis[u] + edge[i].w){ 120 121 int maxf = dfs(v,min(left,edge[i].f)); 122 edge[i].f -= maxf; 123 edge[i^1].f += maxf; 124 left -= maxf; 125 cost += edge[i].w * maxf;//要乘以流量 126 127 if(left <= 0){ 128 left = 0; 129 break; 130 } 131 } 132 } 133 134 return flow - left; 135 } 136 137 void work(){ 138 139 while(SPFA()){ 140 maxflow += dfs(s,maxint); 141 } 142 if(maxflow == N) put(cost); 143 else puts("-1"); 144 } 145 146 int main(){ 147 // file(); 148 readdata(); 149 work(); 150 return 0; 151 }