标签:road constructions hdu 3917 最大权闭合图
4 2 500 10 4 1 2 1 10 2 3 1 20 4 3 1 30 1 4 2 60 4 2 500 100 5 1 2 1 10 2 3 1 20 4 3 1 30 4 3 2 10 1 4 2 60 3 1 10 3 1 2 1 100 2 3 1 100 3 1 1 100 0 0
440 470 0Hintfor second test case, if you choose company 2 responsible ways, then you must choose the path of responsible company 1, but if you choose company 1, then you do not have to choose company 2.
思路:做最大权闭合图的题就是要找到有相互依赖关系的两个集合。题目中告诉了公司之间有依赖关系,那么就以公司为节点,源点和每个公司连边权为纳税,公司和会点连边权为花费,公司与公司之间有联系的连边权为INF。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #pragma comment (linker,"/STACK:102400000,102400000") #define maxn 1005 #define MAXN 5005 #define mod 1000000009 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-6 #define lson rt<<1,l,mid #define rson rt<<1|1,mid+1,r #define FRE(i,a,b) for(i = a; i <= b; i++) #define FRL(i,a,b) for(i = a; i < b; i++) #define mem(t, v) memset ((t) , v, sizeof(t)) #define sf(n) scanf("%d", &n) #define sff(a,b) scanf("%d %d", &a, &b) #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c) #define pf printf #define DBG pf("Hi\n") const int MAXM = 200010; typedef long long ll; using namespace std; struct Edge { int to,next,cap,flow; }edge[MAXM]; int tol; int head[MAXN]; int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN]; void init() { tol=0; memset(head,-1,sizeof(head)); } //加边,单向图三个参数,双向图四个参数 void addedge(int u,int v,int w,int rw=0) { edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u]; edge[tol].flow=0; head[u]=tol++; edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v]; edge[tol].flow=0; head[v]=tol++; } //输入参数:起点,终点,点的总数 //点的编号没有影响,只要输入点的总数 int sap(int start,int end,int N) { memset(gap,0,sizeof(gap)); memset(dep,0,sizeof(dep)); memcpy(cur,head,sizeof(head)); int u=start; pre[u]=-1; gap[0]=N; int ans=0; while (dep[start]<N) { if (u==end) { int Min=INF; for (int i=pre[u];i!=-1;i=pre[edge[i^1].to]) if (Min>edge[i].cap-edge[i].flow) Min=edge[i].cap-edge[i].flow; for (int i=pre[u];i!=-1;i=pre[edge[i^1].to]) { edge[i].flow+=Min; edge[i^1].flow-=Min; } u=start; ans+=Min; continue; } bool flag=false; int v; for (int i=cur[u];i!=-1;i=edge[i].next) { v=edge[i].to; if (edge[i].cap-edge[i].flow && dep[v]+1==dep[u]) { flag=true; cur[u]=pre[v]=i; break; } } if (flag) { u=v; continue; } int Min=N; for (int i=head[u];i!=-1;i=edge[i].next) if (edge[i].cap-edge[i].flow && dep[edge[i].to]<Min) { Min=dep[edge[i].to]; cur[u]=i; } gap[dep[u]]--; if (!gap[dep[u]]) return ans; dep[u]=Min+1; gap[dep[u]]++; if (u!=start) u=edge[pre[u]^1].to; } return ans; } struct Node { int u,v,c,cost; }node[MAXN]; int n,m,num; int cost[MAXN]; int main() { // freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin); int i,j,k,u,v,c,co; while (scanf("%d%d",&n,&m),(n+m)) { init(); int all=0; for (i=1;i<=m;i++) { cost[i]=0; scanf("%d",&c); addedge(0,i,c); all+=c; } scanf("%d",&k); for (i=1;i<=k;i++) { scanf("%d%d%d%d",&node[i].u,&node[i].v,&node[i].c,&node[i].cost); cost[node[i].c]+=node[i].cost; } for (i=1;i<=m;i++) addedge(i,m+1,cost[i]); for (i=1;i<=k;i++) { for (j=1;j<=k;j++) { if (i!=j) { if (node[i].c!=node[j].c&&node[i].v==node[j].u) addedge(node[i].c,node[j].c,INF); } } } printf("%d\n",all-sap(0,m+1,m+2)); } return 0; }
Road constructions (hdu 3917 最大权闭合图)
标签:road constructions hdu 3917 最大权闭合图
原文地址:http://blog.csdn.net/u014422052/article/details/45442607