洛谷P4013 数字梯形问题(费用流)

题意

$N$行的矩阵，第一行有$M$个元素，第$i$行有$M + i - 1$个元素

问在三个规则下怎么取使得权值最大

Sol

我只会第一问qwq。。

因为有数量的限制，考虑拆点建图，把每个点拆为$a_1$和$b_1$，两点之间连流量为$1$，费用为权值的边

从$b_i$向下方和右下的$a_1$连一条流量为$1$，费用为$0$边

从$S$向第一层的$a_1$连流量为$1$，费用为$0$的边，从$b_N$到$T$连流量为$1$，费用为$0$的边

对于第二问，因为没有点的个数的限制，那么就不用拆点了，直接向能到达的点连流量为$1$，费用为点权的边

对于第三问，直接把第二问中的所有边为流量设为$INF$(除了从$S$出发的)

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 1e5 + , INF = 1e9 + ;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = ; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int N, M, S = , T = 1e5 + ;
int a[][];
struct Edge {
    int u, v, w, f, nxt;
}E[MAXN];
int head[MAXN << ], num = ;
inline void add_edge(int x, int y, int w, int f) {

    E[num] = (Edge){x, y, w, f, head[x]};
    head[x] = num++;
}
inline void AddEdge(int x, int y, int w, int f) {
    add_edge(x, y, w, f);
    add_edge(y, x, -w, );
}
int anscost, dis[MAXN], vis[MAXN], Pre[MAXN];
bool SPFA() {
    memset(dis, -0x3f, sizeof(dis));
    memset(vis, , sizeof(vis));
    queue<int> q; q.push(S); dis[S] = ;
    while(!q.empty()) {
        int p = q.front(); q.pop(); vis[p] = ;
        for(int i = head[p]; i !=- ; i = E[i].nxt) {
            int to = E[i].v;
            if((dis[to] < dis[p] + E[i].w) && E[i].f > ) {
                dis[to] = dis[p] + E[i].w;
                Pre[to] = i;
                if(!vis[to]) q.push(to), vis[to] = ;
            }
        }
    }
    return dis[T] > ;
}
int F() {
    int nowflow = INF;
    for(int i = T; i != S; i = E[Pre[i]].u) nowflow = min(nowflow, E[Pre[i]].f);
    for(int i = T; i != S; i = E[Pre[i]].u) E[Pre[i]].f -= nowflow, E[Pre[i] ^ ].f += nowflow;
    anscost += dis[T] * nowflow;
}
int MCMF() {
    anscost = ;
    while(SPFA())
        F();
    return anscost;
}
int be[][], tot = , X;
void Solve1() {
    memset(head, -, sizeof(head)); num = ;
    for(int i = ; i < N; i++) {
        for(int j = ; j <= M + i - ; j++) {
            AddEdge(be[i][j], be[i][j] + X, a[i][j], );
            AddEdge(be[i][j] + X, be[i + ][j], , );
            AddEdge(be[i][j] + X, be[i + ][j + ], , );
        }
    }
    for(int i = ; i <= M; i++) AddEdge(S, be[][i], , );
    for(int i = ; i <= N + M - ; i++)
        AddEdge(be[N][i], be[N][i] + X, a[N][i], ),
        AddEdge(be[N][i] + X, T, , );
    printf("%d\n", MCMF());
}
void Solve2() {
    memset(head, -, sizeof(head)); num = ;
    for(int i = ; i < N; i++) {
        for(int j = ; j <= M + i - ; j++) {
            AddEdge(be[i][j], be[i + ][j + ], a[i][j], );
            AddEdge(be[i][j], be[i + ][j], a[i][j], );
        }
    }
    for(int i = ; i <= M; i++) AddEdge(S, be[][i], , );
    for(int i = ; i <= N + M - ; i++) AddEdge(be[N][i], T, a[N][i], INF);
    printf("%d\n", MCMF());
}
void Solve3() {
    memset(head, -, sizeof(head)); num = ;
    for(int i = ; i < N; i++)
        for(int j = ; j <= M + i - ; j++) {
            AddEdge(be[i][j], be[i + ][j + ], a[i][j], INF);
            AddEdge(be[i][j], be[i + ][j], a[i][j], INF);
        }
    for(int i = ; i <= M; i++) AddEdge(S, be[][i], , );
    for(int i = ; i <= N + M - ; i++) AddEdge(be[N][i], T, a[N][i], INF);
    printf("%d\n", MCMF());
}
int main() {
    memset(head, -, sizeof(head));
    M = read(); N = read(); X = (N + M - ) * N;
    for(int i = ; i <= N; i++)
        for(int j = ; j <= M + i - ; j++)
            a[i][j] = read(), be[i][j] = ++tot;
    Solve1();
    Solve2();
    Solve3();
    return ;
}
/*

*/