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首先最短路长度不同的人肯定不会冲突。
对于最短路长度相同的人,跑个最大流就行了。。当然只有一个人就不用跑了
看起来会T得很惨。。但dinic在单位网络里是O(m*n^0.5)的...
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<algorithm> 5 #include<queue> 6 using namespace std; 7 const int maxn=25233,inf=1002333333; 8 struct zs2{ 9 int too,pre,dis; 10 }e1[100233];int tot1,last1[maxn]; 11 struct zs1{int dis,id;}; 12 priority_queue<zs1>q; 13 bool u[maxn]; 14 int dis1[maxn]; 15 struct zs{ 16 int too,pre;bool flow; 17 }e[533333];int tot,last[maxn]; 18 int dl[maxn]; 19 short dis[maxn]; 20 int pos[1023]; 21 int i,j,k,n,m,ans,s,t,tt,c; 22 23 int ra;char rx; 24 inline int read(){ 25 rx=getchar(),ra=0; 26 while(rx<‘0‘||rx>‘9‘)rx=getchar(); 27 while(rx>=‘0‘&&rx<=‘9‘)ra*=10,ra+=rx-48,rx=getchar();return ra; 28 } 29 30 bool operator <(zs1 a,zs1 b){return a.dis>b.dis;} 31 inline void spfa(){ 32 int i,now; 33 memset(dis1,60,(n+1)<<2),dis1[1]=0,q.push((zs1){0,1}); 34 while(!q.empty()){ 35 while(!q.empty()&&u[q.top().id])q.pop(); 36 if(q.empty())return; 37 now=q.top().id,q.pop(), 38 u[now]=1; 39 for(i=last1[now];i;i=e1[i].pre)if(dis1[e1[i].too]>dis1[now]+e1[i].dis) 40 dis1[e1[i].too]=dis1[now]+e1[i].dis,q.push((zs1){dis1[e1[i].too],e1[i].too}); 41 } 42 } 43 inline void insert1(int a,int b,int c){ 44 e1[++tot1].too=b,e1[tot1].dis=c,e1[tot1].pre=last1[a],last1[a]=tot1, 45 e1[++tot1].too=a,e1[tot1].dis=c,e1[tot1].pre=last1[b],last1[b]=tot1; 46 } 47 bool bfs(){ 48 memset(dis,0,(n+1)<<1); 49 int l=0,r=1,i,now;dl[1]=s,dis[s]=1; 50 while(l<r&&!dis[t]) 51 for(i=last[now=dl[++l]];i;i=e[i].pre)if(e[i].flow&&!dis[e[i].too]) 52 dis[e[i].too]=dis[now]+1,dl[++r]=e[i].too; 53 // for(i=1;i<=n;i++)printf("0->%d %d\n",i,dis[i]); 54 return dis[t]; 55 } 56 int dfs(int x,int mx){ 57 if(x==t)return mx; 58 int used=0,i;bool w; 59 for(i=last[x];i;i=e[i].pre)if(e[i].flow&&dis[e[i].too]==dis[x]+1){ 60 w=dfs(e[i].too,1);if(w){ 61 e[i].flow=0,e[i^1].flow=1,used++; 62 if(used==mx)return mx; 63 } 64 } 65 dis[x]=0;return used; 66 } 67 inline void insert(int a,int b,int c){//printf(" %d-->%d %d\n",a,b,c); 68 e[++tot].too=b,e[tot].flow=c,e[tot].pre=last[a],last[a]=tot; 69 e[++tot].too=a,e[tot].flow=0,e[tot].pre=last[b],last[b]=tot; 70 } 71 inline int check(int L,int R){//printf("check: %d--%d\n",L,R); 72 register int i;int flow=0,j; 73 for(i=2;i<=tt;i+=2)e[i].flow=1,e[i^1].flow=0; 74 for(j=1;i<=tot;i+=2,j++) 75 e[i].flow=j>=L&&j<=R,e[i^1].flow=0; 76 while(bfs())flow+=dfs(s,inf); 77 // printf("flow: %d\n",flow); 78 return flow; 79 } 80 bool cmp(int a,int b){return dis1[a]<dis1[b];} 81 int main(){ 82 n=read(),m=read(),c=read(); 83 for(i=1;i<=m;i++)j=read(),k=read(),insert1(j,k,read()); 84 for(i=1;i<=c;i++)pos[i]=read(); 85 spfa(); 86 // for(i=1;i<=n;i++)printf("1-->%d %d\n",i,dis1[i]); 87 s=0,t=1,tot=1; 88 for(i=1;i<=n;i++)for(j=last1[i];j;j=e1[j].pre) 89 if(dis1[e1[j].too]==dis1[i]+e1[j].dis)insert(e1[j].too,i,1);tt=tot; 90 sort(pos+1,pos+1+c,cmp); 91 92 for(i=1;i<=c;i++)insert(s,pos[i],0); 93 int pre; 94 for(i=1;i<=c&&pos[i]==1;i++,ans++); 95 for(pre=i;i<=c;i++)if(dis1[pos[i]]!=dis1[pos[i+1]]||i==c){ 96 if(pre==i)ans++;else ans+=check(pre,i); 97 pre=i+1; 98 } 99 printf("%d\n",ans); 100 return 0; 101 }
dinic好优越啊...
[bzoj3955] [WF2013]Surely You Congest
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原文地址:http://www.cnblogs.com/czllgzmzl/p/5597046.html