CSU 1506 - Double Shortest Paths(网络流’最小费用流)

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题目：

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=75853#problem/D

题意：

求出两个人从1走到n的最小花费，前一个人走过的路径权值会改变为d+a。

思路：

两点之间建两条边，分别为容量1费用d 和容量1费用d+a。由于模板求的是最小费用最大流，而本题需要限制流量为2，所以需要另设源点和汇点(容量为2) 来限制流量.

AC.

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 505;const int maxm = 2005;struct Edge {    int to, next, cap, flow, cost;}edge[maxm];int head[maxn], tol;int pre[maxn], dis[maxn];bool vis[maxn];int N;void init(int n){    N = n;    tol = 0;    memset(head, -1, sizeof(head));    //memset(pre, -1, sizeof(pre));}void addedge(int u, int v, int cap, int cost){    edge[tol].to = v;    edge[tol].cap = cap;    edge[tol].cost = cost;    edge[tol].flow = 0;    edge[tol].next = head[u];    head[u] = tol++;    edge[tol].to = u;    edge[tol].cap = 0;    edge[tol].cost = -cost;    edge[tol].flow = 0;    edge[tol].next = head[v];    head[v] = tol++;}bool spfa(int s, int t){    queue<int> q;    for(int i = 0; i < N; ++i) {        dis[i] = INF;        vis[i] = false;        pre[i] = -1;    }    dis[s] = 0;    vis[s] = true;    q.push(s);    while(!q.empty()) {        int u = q.front(); q.pop();        vis[u] = false;        for(int i = head[u]; i != -1; i = edge[i].next) {            int v = edge[i].to;            if(edge[i].cap > edge[i].flow &&                dis[v] > dis[u] + edge[i].cost) {                dis[v] = dis[u] + edge[i].cost;                pre[v] = i;                if(!vis[v]) {                    vis[v] = true;                    q.push(v);                }            }        }    }    if(pre[t] == -1) return false;    else return true;}int mincostmaxflow(int s, int t, int &cost){    int flow = 0;    cost = 0;    while(spfa(s, t)) {        int Min = INF;        for(int i = pre[t]; 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[t]; i != -1; i = pre[edge[i^1].to]) {            edge[i].flow += Min;            edge[i^1].flow -= Min;            cost += edge[i].cost * Min;        }//printf("%d\n", flow);        flow += Min;    }    return flow;}int main(){    //freopen("in", "r", stdin);    int ca = 1;    int n, m;    while(~scanf("%d %d", &n, &m)) {        int s = 0, t = n + 1;        int ans;        init(t+1); //        for(int i = 0; i < m; ++i) {            int u, v, c, a;            scanf("%d %d %d %d", &u, &v, &c, &a);            addedge(u, v, 1, c);            addedge(u, v, 1, c+a);        }        addedge(s, 1, 2, 0);        addedge(n, t, 2, 0);        mincostmaxflow(s, t, ans);        printf("Case %d: %d\n", ca++, ans);    }    return 0;}

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