网站图片链接到视频怎么做,西宁集团网站建设,软件网站排名,揭阳做网站最大流问题。ISAP算法。注意可能会有重边#xff0c;不过我用的数据结构支持重边。距离d我直接初始化为0#xff0c;也可以用BFS逆向找一次。-----------------------------------------------------------------------#includecstdio#includeiostream#inclu… 最大流问题。ISAP算法。注意可能会有重边不过我用的数据结构支持重边。距离d我直接初始化为0也可以用BFS逆向找一次。-----------------------------------------------------------------------#includecstdio#includeiostream#includealgorithm#includecstring#includevector#includequeue#define rep(i,l,r) for(int il;ir;i)#define dow(i,l,r) for(int il;ir;i--)#define clr(x,c) memset(x,c,sizeof(x))using namespace std;const int inf0x3f3f3f3f,maxn2005;struct edge { int from,to,cap,flow;};struct ISAP { int n,m,s,t; vectoredge edges; vectorint g[maxn]; int d[maxn]; int cur[maxn]; int p[maxn]; int num[maxn]; void init(int n) { this-nn; rep(i,0,n) g[i].clear(); edges.clear(); } void addEdge(int from,int to,int cap) { edges.push_back((edge){from,to,cap,0}); edges.push_back((edge){to,from,0,0}); medges.size(); g[from].push_back(m-2); g[to].push_back(m-1); } int augment() { int xt,ainf; while(x!s) { edge eedges[p[x]]; amin(a,e.cap-e.flow); xedges[p[x]].from; } xt; while(x!s) { edges[p[x]].flowa; edges[p[x]^1].flow-a; xedges[p[x]].from; } return a; } int maxFlow(int s,int t) { this-ss; this-tt; int flow0; clr(d,0); clr(num,0); rep(i,0,n) num[d[i]]; int xs; clr(cur,0); while(d[s]n) { if(xt) { flowaugment(); xs; } int ok0; rep(i,cur[x],g[x].size()) { edge eedges[g[x][i]]; if(e.cape.flow d[x]d[e.to]1) { ok1; p[e.to]g[x][i]; cur[x]i; xe.to; break; } } if(!ok) { int mn-1; rep(i,0,g[x].size()) { edge eedges[g[x][i]]; if(e.cape.flow) mmin(m,d[e.to]); } if(--num[d[x]]0) break; num[d[x]m1]; cur[x]0; if(x!s) xedges[p[x]].from; } } return flow; }} isap;int s() { int n,m; cinmn; isap.init(n); rep(i,0,m) { int from,to,cap,pd1; scanf(%d%d%d,from,to,cap); isap.addEdge(from-1,to-1,cap); } return isap.maxFlow(0,n-1);}int main() { freopen(ditch.in,r,stdin); freopen(ditch.out,w,stdout); couts()endl; return 0;}----------------------------------------------------------------------- Drainage DitchesHal Burch Every time it rains on Farmer Johns fields, a pond forms over Bessies favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessies clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch. Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network. Note however, that there can be more than one ditch between two intersections. Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle. PROGRAM NAME: ditch INPUT FORMAT Line 1:Two space-separated integers, N (0 N 200) and M (2 M 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream.Line 2..N1:Each of N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 Si, Ei M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 Ci 10,000,000) is the maximum rate at which water will flow through the ditch.SAMPLE INPUT (file ditch.in) 5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10OUTPUT FORMAT One line with a single integer, the maximum rate at which water may emptied from the pond. SAMPLE OUTPUT (file ditch.out) 50转载于:https://www.cnblogs.com/JSZX11556/p/4295472.html
