HDU 5889 最短路 网络流

题目

Barricade

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1702    Accepted Submission(s): 502

Problem Description

The empire is under attack again. The general of empire is planning to defend his castle. The land can be seen as N towns and M roads, and each road has the same length and connects two towns. The town numbered 1 is where general's castle is located, and the town numbered N is where the enemies are staying. The general supposes that the enemies would choose a shortest path. He knows his army is not ready to fight and he needs more time. Consequently he decides to put some barricades on some roads to slow down his enemies. Now, he asks you to find a way to set these barricades to make sure the enemies would meet at least one of them. Moreover, the barricade on the i-th road requires wi units of wood. Because of lacking resources, you need to use as less wood as possible.

 

Input

The first line of input contains an integer t, then t test cases follow.
For each test case, in the first line there are two integers N(N≤1000) and M(M≤10000).
The i-the line of the next M lines describes the i-th edge with three integers u,v and w where 0≤w≤1000 denoting an edge between u and v of barricade cost w.

 

Output

For each test cases, output the minimum wood cost.

 

Sample Input

14 41 2 12 4 23 1 34 3 4

 

Sample Output

4

题目大意

   通俗的讲就是给了你一堆路，让你求出从1->n的最短路，然后在最短路中求一个最大流，最大流自己不会，这里直接套的模板 （Dinic）

解题思路如题意

#include<queue>#include<cstdio>#include<vector>#include<cstring>#include<iostream>using namespace std;const int INF=0x3f3f3f3f;int n;int d[1000+10];int vis[1000+10];struct point{    int v;    int dis;    point(){}    point (int vv,int d):v(vv),dis(d){}};vector<point> mye[1000+10];void spfa(int x) {     d[x]=0;     vis[x]=1;     queue<int>qu;     qu.push(x);     while(!qu.empty())     {         int u=qu.front();         qu.pop();         for(auto &vb:mye[u])         {             if(d[u]+1<=d[vb.v])             {                 d[vb.v]=d[u]+1;                 if(!vis[vb.v])                 {                     vis[vb.v]=1;                     qu.push(vb.v);                 }             }         }         vis[u]=0;     } }#define M 10000+10struct node {    int u, v, next, cap;} edge[M*5];int nex[M], head[M], layer[M], Q[M * 2], mark[M];int ecnt;void init(){    ecnt= 0;    memset(head,-1,sizeof(head));}void add(int u, int v, int c) {    edge[ecnt].u = u;    edge[ecnt].v = v;    edge[ecnt].cap = c;    edge[ecnt].next = head[u];    head[u] = ecnt++;    edge[ecnt].u = v;    edge[ecnt].v = u;    edge[ecnt].cap = 0;    edge[ecnt].next = head[v];    head[v] = ecnt++;}bool BFS(int begin, int end) {    int i, l, h, k, y;    for (i = 0; i <= end; i++) layer[i] = -1;    layer[begin] = 0;    l = h = 0;    Q[l++] = begin;    while (h < l) {        k = Q[h++];        for (i = head[k]; i != -1; i = edge[i].next) {            y = edge[i].v;            if (edge[i].cap > 0 && layer[y] == -1) {                layer[y] = layer[k] + 1;                if (y == end)                    return true;                Q[l++] = y;            }        }    }    return false;}int DFS(int x, int exp, int end) {    mark[x] = 1;    if (x == end)return exp;    int y, temp, i;    for (i = nex[x]; i != -1; i = edge[i].next, nex[x] = i) {        y = edge[i].v;        if (edge[i].cap > 0 && layer[y] == layer[x] + 1 && !mark[y]) {            if ((temp = (DFS(y, min(exp, edge[i].cap), end))) > 0) {                edge[i].cap -= temp;//流完后正向流表示剩余流量                edge[i^1].cap += temp;//流完后反向流表示正向流的流量                return temp;            }        }    }    return 0;}int Dinic_flow(int begin, int end) {    int i, ans = 0, flow;    while (BFS(begin, end)) {        for (i = 0; i <= end; i++)nex[i] = head[i];        while (true) {            for (i = 0; i <= end; i++) mark[i] = 0;            flow = DFS(begin, INT_MAX, end);            if (flow == 0)break;            ans += flow;        }    }    return ans;} void work() {     int n,m;     init();     scanf("%d%d",&n,&m);     for(int i=0;i<=1000;i++)        mye[i].clear();     int x,y,z;     for(int i=0;i<m;i++)        {            scanf("%d %d %d",&x,&y,&z);            mye[x].push_back(point(y,z));            mye[y].push_back(point(x,z));        }     memset(d,INF,sizeof(d));     memset(vis,0,sizeof(vis));     spfa(1);     for(int i=1;i<=n;i++)     {         for(auto &y:mye[i])         {             if(d[i]+1==d[y.v])             {                add(i,y.v,y.dis);             }         }     }     int ans=Dinic_flow(1,n);     printf("%d\n",ans); } int main() {     int T;     scanf("%d",&T);     while(T--)     {         work();     }     return 0;}

ISAP做法

#include<bits/stdc++.h>using namespace std;const int MAXN = 100010;//点数的最大值const int MAXM = 400010;//边数的最大值const int INF = 0x3f3f3f3f;int d[MAXN] , vis[MAXN];struct point{    int to ;    int dis ;};vector<point> e[MAXN] ;void spfa(int s){    for(int i = 0 ; i < MAXN ; i++)        d[i] = INF ;    d[s] = 0 ;    queue<int> Q ;    vis[s] = 1 ;    Q.push(s) ;    while(!Q.empty())    {        int t = Q.front() ;        Q.pop() ;        for(auto &v : e[t])        {            if(d[v.to] > d[t] + 1)            {                d[v.to] = d[t] + 1 ;                vis[v.to] = 1 ;                Q.push(v.to) ;            }        }        vis[t] = 0 ;    }}struct Edge{    int to,next,cap,flow;} edge[MAXM]; //注意是MAXMint 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;}int main(){    int t ;    scanf("%d" , &t) ;    int n,m;    while(t--)    {        scanf("%d%d",&n,&m) ;        int u,v,w,sum=0;        init();        for(int i = 0 ; i <= n ; i++) e[i].clear() ;        for(int i=0; i<m; i++)        {            scanf("%d%d%d",&u,&v,&w);            e[u].push_back({v,w}) ;            e[v].push_back({u,w}) ;        }        spfa(1) ;        for(int i = 1 ; i <= n ; i++)        {            for(auto &y : e[i])            {                if(d[i] + 1 == d[y.to])                 {                     addedge(i,y.to,y.dis);                  //   cout<<i<<"  " << y.to <<endl ;                 }            }        }        printf("%d\n",sap(1,n,n));    }    return 0;}