【题目链接】
http://www.lydsy.com/JudgeOnline/problem.php?id=3931
【题意】
只能通过1-n的最短路,求网络最大流
【思路】
分别以1,n为起点做最短路,则可以判断一条边是否在最短路上。
以最短路构建网络,并且将一个点拆成两个中间连c[i]表示结点容量限制。
【代码】
1 #include<set> 2 #include<cmath> 3 #include<queue> 4 #include<vector> 5 #include<cstdio> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 #define trav(u,i) for(int i=front[u];i;i=e[i].nxt) 10 #define FOR(a,b,c) for(int a=(b);a<=(c);a++) 11 using namespace std; 12 13 typedef long long ll; 14 const int N = 2e3+10; 15 const int M = 4e5+10; 16 const int inf = 1e9; 17 const ll llinf = 1e18; 18 19 ll read() { 20 char c=getchar(); 21 ll f=1,x=0; 22 while(!isdigit(c)) { 23 if(c=='-') f=-1; c=getchar(); 24 } 25 while(isdigit(c)) 26 x=x*10+c-'0',c=getchar(); 27 return x*f; 28 } 29 30 struct DEdge { 31 int u,v; ll cap,flow; 32 }; 33 struct Dinic { 34 int n,m,s,t; 35 int d[N],cur[N],vis[N]; 36 vector<int> g[N]; 37 vector<DEdge> es; 38 queue<int> q; 39 void init(int n) { 40 this->n=n; 41 es.clear(); 42 FOR(i,0,n) g[i].clear(); 43 } 44 void clear() { 45 FOR(i,0,(int)es.size()-1) es[i].flow=0; 46 } 47 void AddEdge(int u,int v,ll w) { 48 es.push_back((DEdge){u,v,w,0}); 49 es.push_back((DEdge){v,u,0,0}); 50 m=es.size(); 51 g[u].push_back(m-2); 52 g[v].push_back(m-1); 53 } 54 int bfs() { 55 memset(vis,0,sizeof(vis)); 56 q.push(s); d[s]=0; vis[s]=1; 57 while(!q.empty()) { 58 int u=q.front(); q.pop(); 59 FOR(i,0,(int)g[u].size()-1) { 60 DEdge& e=es[g[u][i]]; 61 int v=e.v; 62 if(!vis[v]&&e.cap>e.flow) { 63 vis[v]=1; 64 d[v]=d[u]+1; 65 q.push(v); 66 } 67 } 68 } 69 return vis[t]; 70 } 71 ll dfs(int u,ll a) { 72 if(u==t||!a) return a; 73 ll flow=0,f; 74 for(int& i=cur[u];i<g[u].size();i++) { 75 DEdge& e=es[g[u][i]]; 76 int v=e.v; 77 if(d[v]==d[u]+1&&(f=dfs(v,min(a,e.cap-e.flow)))>0) { 78 e.flow+=f; 79 es[g[u][i]^1].flow-=f; 80 flow+=f; a-=f; 81 if(!a) break; 82 } 83 } 84 return flow; 85 } 86 ll MaxFlow(int s,int t) { 87 this->s=s,this->t=t; 88 ll flow=0; 89 while(bfs()) { 90 memset(cur,0,sizeof(cur)); 91 flow+=dfs(s,llinf); 92 } 93 return flow; 94 } 95 } dc; 96 97 struct Edge { 98 int u,v;ll w; int nxt; 99 }e[M]; 100 int en=1,front[N]; 101 void adde(int u,int v,ll w) 102 { 103 e[++en]=(Edge){u,v,w,front[u]}; front[u]=en; 104 } 105 106 int n,m; 107 ll c[N],dis1[N],dis2[N]; 108 109 int inq[N]; queue<int> q; 110 void spfa(int s,ll* dis) 111 { 112 memset(inq,0,sizeof(inq)); 113 FOR(i,0,n) dis[i]=llinf; 114 q.push(s); inq[s]=1; dis[s]=0; 115 while(!q.empty()) { 116 int u=q.front(); q.pop(); 117 inq[u]=0; 118 trav(u,i) { 119 int v=e[i].v; 120 if(dis[v]>dis[u]+(ll)e[i].w) { 121 dis[v]=dis[u]+(ll)e[i].w; 122 if(!inq[v]) { 123 inq[v]=1; q.push(v); 124 } 125 } 126 } 127 } 128 } 129 130 int main() 131 { 132 n=read(),m=read(); 133 FOR(i,1,m) { 134 int u=read(),v=read(); ll w=read(); 135 adde(u,v,w),adde(v,u,w); 136 } 137 dc.init(n*2+2); 138 spfa(1,dis1),spfa(n,dis2); 139 ll dist=dis1[n]; 140 FOR(i,1,n) 141 c[i]=read(),dc.AddEdge(i,i+n,c[i]); 142 for(int i=2;i<=en;i+=2) { 143 int u=e[i].u,v=e[i].v; 144 if(dis1[u]>dis1[v]) swap(u,v); 145 if((ll)dis1[u]+e[i].w+dis2[v]==dist) 146 dc.AddEdge(u+n,v,inf),dc.AddEdge(v+n,u,inf); 147 } 148 printf("%lld\n",dc.MaxFlow(n+1,n)); 149 return 0; 150 }
P.S. spfa写错还浑然不知,省选药丸的节奏QWQ