題意：

給一個網(wǎng)絡(luò)中某些邊增加容量，增加的總和最大為K，使得最大流最大。

費(fèi)用流:在某條邊增加單位流量的費(fèi)用。

那么就可以2個點(diǎn)之間建2條邊，第一條給定邊(u,v,x,0)這條邊費(fèi)用為0

同時另一條邊(u,v,K,1)費(fèi)用為1，那么就可以通過限制在增廣時相應(yīng)的費(fèi)用即可找出最大流

個人覺得這樣做的原因是每次增光都是最優(yōu)的。所以通過限制最終費(fèi)用不超過K可以得到最優(yōu)解

#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 110; const int INF = 0x3f3f3f3f; struct node {int u,v,next;int flow,cap,cost; }edge[MAXN * MAXN * 4]; int cnt,src,tag; int C,F; int K,N; queue<int>q; bool inq[MAXN];int d[MAXN]; int head[MAXN],p[MAXN]; int tot = 0;void init() {memset(head,-1,sizeof(head));tot = 0; }void add_edge(int u,int v,int cap,int cost) {edge[cnt].u = u;edge[cnt].v = v;edge[cnt].cap = cap;edge[cnt].flow = 0;edge[cnt].cost = cost;edge[cnt].next = head[u];head[u] = cnt++;//反向edge[cnt].v = u;edge[cnt].u = v;edge[cnt].flow = 0;edge[cnt].cap = 0;edge[cnt].cost = - cost;edge[cnt].next = head[v];head[v] = cnt++; }bool SPFA(int s, int t) {while (!q.empty()) q.pop();memset(inq,false,sizeof(inq));memset(d,0x3f,sizeof(d));memset(p,-1,sizeof(p));d[s] = 0;q.push(s);inq[s] = true;while (!q.empty()){int u = q.front(); q.pop();inq[u] = false;for (int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if (d[v] > d[u] + edge[i].cost && edge[i].cap > edge[i].flow){d[v] = d[u] + edge[i].cost;p[v] = i;if (!inq[v]){q.push(v);inq[v] = true;}}}}if(d[tag] == INF) return false;int a = INF;for (int i = p[tag]; i != -1; i = p[edge[i].u])a = min(a,edge[i].cap - edge[i].flow);if(C + d[tag] * a > K){F += (K - C) / d[tag];return false;}return true; } void slove() {C = F = 0;while(SPFA(src,tag)){int a = INF;for (int i = p[tag]; i != -1; i = p[edge[i].u])a = min(a,edge[i].cap - edge[i].flow);for (int i = p[tag]; i != -1; i = p[edge[i].u]){edge[i].flow += a;edge[i ^ 1].flow -= a;}C += d[tag] * a;F += a;} }int main() {while (scanf("%d%d",&N,&K) != EOF){init();for (int i = 1 ; i <= N ; i++)for (int j = 1 ; j <= N ; j++){int x;scanf("%d",&x);if (x){add_edge(i,j,x,0);add_edge(i,j,K,1);}}src = 1;tag = N;slove();printf("%d\n",F);}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/Commence/p/4928476.html

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