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题解:
总算A掉了,各种蛋疼。。。
int main() { freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); n=read();m=read();s=0;t=2*n+m+1;ss=t+1;tt=t+2;sss=tt+1; insert(s,sss,0,m,0); for1(i,n)insert(sss,i,0,inf,0); for1(i,n){int x=read();insert(i,i+n,x,x,0);} for1(i,n)for2(j,i+1,n){int x=read();if(x!=-1)insert(i+n,j,0,inf,x);} for1(i,n)insert(i+n,t,0,inf,0); insert(t,s,0,inf,0); mcf(); printf("%d\n",mincost); return 0; }
s是附加源,sss是真正的源,t是真正的汇。这样构图就好了,但我们还有下界限制,那再来个ss,tt。。。居然还是费用流。。。然后这题就做完了
因为我们只需要最小费用,所以我们只需求出可行流&最小费用即可。
关于下界的处理我直接写在insert函数里。。。导致慢成翔。。。
代码:
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set> 10 #include<queue> 11 #include<string> 12 #define inf 1000000000 13 #define maxn 200000 14 #define maxm 200000 15 #define eps 1e-10 16 #define ll long long 17 #define pa pair<int,int> 18 #define for0(i,n) for(int i=0;i<=(n);i++) 19 #define for1(i,n) for(int i=1;i<=(n);i++) 20 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 21 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 22 #define for4(i,x) for(int i=head[x],y;i;i=e[i].next) 23 #define mod 1000000007 24 using namespace std; 25 inline int read() 26 { 27 int x=0,f=1;char ch=getchar(); 28 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 29 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();} 30 return x*f; 31 } 32 int n,m,k,mincost,tot=1,s,t,ss,tt,sss,head[maxn],d[maxn],from[2*maxm]; 33 bool v[maxn]; 34 queue<int>q; 35 struct edge{int from,go,next,v,c;}e[2*maxm]; 36 void ins(int x,int y,int v,int c) 37 { 38 e[++tot]=(edge){x,y,head[x],v,c};head[x]=tot; 39 e[++tot]=(edge){y,x,head[y],0,-c};head[y]=tot; 40 } 41 void insert(int x,int y,int l,int r,int c) 42 { 43 ins(ss,y,l,c);ins(x,tt,l,0);ins(x,y,r-l,c); 44 } 45 bool spfa() 46 { 47 for (int i=0;i<=sss;i++){v[i]=0;d[i]=inf;} 48 q.push(ss);d[ss]=0;v[ss]=1; 49 while(!q.empty()) 50 { 51 int x=q.front();q.pop();v[x]=0; 52 for (int i=head[x],y;i;i=e[i].next) 53 if(e[i].v&&d[x]+e[i].c<d[y=e[i].go]) 54 { 55 d[y]=d[x]+e[i].c;from[y]=i; 56 if(!v[y]){v[y]=1;q.push(y);} 57 } 58 } 59 return d[tt]!=inf; 60 } 61 void mcf() 62 { 63 mincost=0; 64 while(spfa()) 65 { 66 int tmp=inf; 67 for(int i=from[tt];i;i=from[e[i].from]) tmp=min(tmp,e[i].v); 68 mincost+=d[tt]*tmp; 69 for(int i=from[tt];i;i=from[e[i].from]){e[i].v-=tmp;e[i^1].v+=tmp;} 70 } 71 } 72 int main() 73 { 74 freopen("input.txt","r",stdin); 75 freopen("output.txt","w",stdout); 76 n=read();m=read();s=0;t=2*n+m+1;ss=t+1;tt=t+2;sss=tt+1; 77 insert(s,sss,0,m,0); 78 for1(i,n)insert(sss,i,0,inf,0); 79 for1(i,n){int x=read();insert(i,i+n,x,x,0);} 80 for1(i,n)for2(j,i+1,n){int x=read();if(x!=-1)insert(i+n,j,0,inf,x);} 81 for1(i,n)insert(i+n,t,0,inf,0); 82 insert(t,s,0,inf,0); 83 mcf(); 84 printf("%d\n",mincost); 85 return 0; 86 }
UPD:把边合并了之后用时成了1/4
代码:
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set> 10 #include<queue> 11 #include<string> 12 #define inf 1000000000 13 #define maxn 200000 14 #define maxm 200000 15 #define eps 1e-10 16 #define ll long long 17 #define pa pair<int,int> 18 #define for0(i,n) for(int i=0;i<=(n);i++) 19 #define for1(i,n) for(int i=1;i<=(n);i++) 20 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 21 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 22 #define for4(i,x) for(int i=head[x],y;i;i=e[i].next) 23 #define mod 1000000007 24 using namespace std; 25 inline int read() 26 { 27 int x=0,f=1;char ch=getchar(); 28 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 29 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();} 30 return x*f; 31 } 32 int n,m,k,mincost,tot=1,s,t,ss,tt,sss,head[maxn],d[maxn],from[2*maxm],in[maxn]; 33 bool v[maxn]; 34 queue<int>q; 35 struct edge{int from,go,next,v,c;}e[2*maxm]; 36 void ins(int x,int y,int v,int c) 37 { 38 e[++tot]=(edge){x,y,head[x],v,c};head[x]=tot; 39 e[++tot]=(edge){y,x,head[y],0,-c};head[y]=tot; 40 } 41 void insert(int x,int y,int l,int r,int c) 42 { 43 in[y]+=l; 44 in[x]-=l; 45 ins(x,y,r-l,c); 46 } 47 void build() 48 { 49 for0(i,sss)if(in[i]>0)ins(ss,i,in[i],0); 50 else if(in[i]<0)ins(i,tt,-in[i],0); 51 } 52 bool spfa() 53 { 54 for (int i=0;i<=sss;i++){v[i]=0;d[i]=inf;} 55 q.push(ss);d[ss]=0;v[ss]=1; 56 while(!q.empty()) 57 { 58 int x=q.front();q.pop();v[x]=0; 59 for (int i=head[x],y;i;i=e[i].next) 60 if(e[i].v&&d[x]+e[i].c<d[y=e[i].go]) 61 { 62 d[y]=d[x]+e[i].c;from[y]=i; 63 if(!v[y]){v[y]=1;q.push(y);} 64 } 65 } 66 return d[tt]!=inf; 67 } 68 void mcf() 69 { 70 mincost=0; 71 while(spfa()) 72 { 73 int tmp=inf; 74 for(int i=from[tt];i;i=from[e[i].from]) tmp=min(tmp,e[i].v); 75 mincost+=d[tt]*tmp; 76 for(int i=from[tt];i;i=from[e[i].from]){e[i].v-=tmp;e[i^1].v+=tmp;} 77 } 78 } 79 int main() 80 { 81 freopen("input.txt","r",stdin); 82 freopen("output.txt","w",stdout); 83 n=read();m=read();s=0;t=2*n+m+1;ss=t+1;tt=t+2;sss=tt+1; 84 insert(s,sss,0,m,0); 85 for1(i,n)insert(sss,i,0,inf,0); 86 for1(i,n){int x=read();insert(i,i+n,x,x,0);} 87 for1(i,n)for2(j,i+1,n){int x=read();if(x!=-1)insert(i+n,j,0,inf,x);} 88 for1(i,n)insert(i+n,t,0,inf,0); 89 insert(t,s,0,inf,0); 90 build(); 91 mcf(); 92 printf("%d\n",mincost); 93 return 0; 94 }
第一行两个正整数N,M。第二行有N个不大于M正整数,分别表示V1,V2......VN。接下来有N ¡ 1行。第i行有N ¡ i个整数,该行的第j个数表示从第i个国家到第i + j个国家的机票费(如果该值等于¡1则表示这两个国家间没有通航)。
在第一行输出最少的总费用。
1<= N < =100 1<= M <= 79
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原文地址:http://www.cnblogs.com/zyfzyf/p/4176415.html