這道題么= =還是有些惡心的，第一次寫帶上下界的網絡流，整個人都萌萌噠~~~

首先先預處理得最短路后

直接用費用流做就行了。

第一次寫，還是挺好寫的= =

CODE：

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<queue>

using namespace std;

#define maxn 310

#define maxm 100000

struct edges{

????int to,next,cap,dist;

}edge[maxm];

int next[maxn],l;

void addedge(int x,int y,int z,int d){

????l++;

????edge[l*2]=(edges){y,next[x],z,d};next[x]=l*2;

????edge[l*2+1]=(edges){x,next[y],0,-d};next[y]=l*2+1;

}

int inf;

int dist[maxn],w[maxn],s,t,cnt;

bool b[maxn];

queue<int> q;

bool spfa() {

????for (int i=1;i<=cnt;i++) dist[i]=inf;

????dist[s]=0;

????q.push(s);

????while (!q.empty()) {

????????int u=q.front();q.pop();

????????b[u]=0;

????????for (int i=next[u];i;i=edge[i].next)

????????????if (edge[i].cap&&edge[i].dist+dist[u]<dist[edge[i].to]) {

????????????????dist[edge[i].to]=dist[u]+edge[i].dist;

????????????????w[edge[i].to]=i;

????????????????if (!b[edge[i].to]) {

????????????????????b[edge[i].to]=1;

????????????????????q.push(edge[i].to);

????????????????}

????????????}

????}

????return dist[t]!=inf;

}

int ans;

int mcmf(int ss,int tt){

????s=ss,t=tt;

????while (spfa()) {

????????int x=t,flow=inf;

????????while (x!=s) {

????????????flow=min(flow,edge[w[x]].cap);

????????????x=edge[w[x]^1].to;

????????}

????????ans+=flow*dist[t];

????????x=t;

????????while (x!=s) {

????????????edge[w[x]].cap-=flow;

????????????edge[w[x]^1].cap+=flow;

????????????x=edge[w[x]^1].to;

????????}

????}

}

int dis[maxn][maxn],f[maxn][maxn],id[maxn][2];

int main(){

????int n,m,k;

????scanf("%d%d%d",&n,&m,&k);

????memset(f,10,sizeof(f));

????inf=f[0][0];

????for (int i=1;i<=m;i++) {

????????int x,y,z;

????????scanf("%d%d%d",&x,&y,&z);

????????f[x][y]=f[y][x]=min(f[x][y],z);

????}

????for (int i=0;i<=n;i++) f[i][i]=0;

????for (int k=0;k<=n;k++) {

????????for (int i=0;i<=n;i++)

????????????????for (int j=0;j<=n;j++)

????????????????????f[i][j]=min(f[i][j],f[i][k]+f[k][j]);

????????for (int i=0;i<=n;i++) dis[i][k]=f[i][k];

????}

????for (int i=0;i<=n;i++) id[i][0]=++cnt;

????for (int i=0;i<=n;i++) id[i][1]=++cnt;

????int s=++cnt,t=++cnt;

????addedge(id[0][0],id[0][1],k,0);

????for (int i=1;i<=n;i++) {

????????addedge(id[0][1],id[i][0],inf,dis[0][i]);

????????addedge(id[i][1],id[n][1],inf,0);

????????addedge(id[i][0],id[i][1],inf,0);

????????addedge(s,id[i][1],1,0);

????????addedge(id[i][0],t,1,0);

????}

????for (int i=0;i<n;i++)

????????for (int j=i+1;j<=n;j++)

???????????if (dis[i][j]!=inf) addedge(id[i][1],id[j][0],inf,dis[i][j]);

????addedge(id[n][1],id[0][0],inf,0);

????mcmf(s,t);

????next[s]=next[t]=0;

????next[id[n][1]]=edge[next[id[n][1]]].next;

????next[id[0][0]]=edge[next[id[0][0]]].next;

????mcmf(id[0][0],id[n][1]);

????printf("%d\n",ans);

????return 0;

}

轉載于:https://www.cnblogs.com/New-Godess/p/4348905.html

總結

以上是生活随笔為你收集整理的BZOJ 2324: [ZJOI2011]营救皮卡丘(带上下限的最小费用最大流)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。