poj2396 Budget 上下界可行流
阿新 • • 發佈：2019-03-16
大根堆 || () 黃金 dde sum iter lin targe 
Budget：http://poj.org/problem?id=2396
題意：
給定一個棋盤，給定每一行每一列的和，還有每個點的性質。求一個合理的棋盤數值放置方式。
思路：
比較經典的網絡流模型，把每一列看成一個點，每一行看成一個點，利用上下界可行流的思路建圖就行了，註意這裏由於是嚴格的小於和大於，所以可以利用 x+1, x-1。
還有就是這道題的0 ， 0 說的是對整張圖的操作。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #includeView Code<cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include<map> #include <set> #include <cassert> /* ⊂_ヽ ＼＼ Λ＿Λ 來了老弟 ＼(‘?‘) > ⌒ヽ / へ＼ / / ＼＼ ? ノ ヽ_つ / / / /| ( (ヽ | |、＼ | 丿 ＼ ⌒) | | ) / ‘ノ ) L? */ using namespace std; #define lson (l , mid , rt << 1) #definerson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//這是一個大根堆q //priority_queue<int,vector<int>,greater<int> >q;//這是一個小根堆q #define fi first #define se second //#define endl ‘\n‘ #define boost ios::sync_with_stdio(false);cin.tie(0) #define rep(a, b, c) for(int a = (b); a <= (c); ++ a) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黃金分割點 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar(); while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar(); return x=f?-x:x; } inline void cmax(int &x,int y){if(x<y)x=y;} inline void cmax(ll &x,ll y){if(x<y)x=y;} inline void cmin(int &x,int y){if(x>y)x=y;} inline void cmin(ll &x,ll y){if(x>y)x=y;} /*-----------------------showtime----------------------*/ const int maxn = 1e4+9; int n,m; struct E{ int v,val,id; int nxt; }edge[maxn]; int head[maxn],gtot; void addedge(int u,int v,int val, int id){ edge[gtot].v = v; edge[gtot].val = val; edge[gtot].nxt = head[u]; edge[gtot].id = -1; head[u] = gtot++; edge[gtot].v = u; edge[gtot].val = 0; edge[gtot].nxt = head[v]; edge[gtot].id = id; head[v] = gtot++; } int dis[maxn],cur[maxn],all; bool bfs(int s,int t) { memset(dis, inf, sizeof(dis)); for(int i=0; i<=all; i++) cur[i] = head[i]; queue<int>que; que.push(s); dis[s] = 0; while(!que.empty()){ int u = que.front(); que.pop(); for(int i=head[u]; ~i; i=edge[i].nxt){ int v = edge[i].v,val = edge[i].val; if(val > 0 && dis[v] > dis[u] + 1){ dis[v] = dis[u]+ 1; que.push(v); } } } return dis[t] < inf; } int dfs(int u,int t,int maxflow){ if(u == t || maxflow == 0) return maxflow; for(int i=cur[u]; ~i; i=edge[i].nxt){ cur[u] = i; int v=edge[i].v,val = edge[i].val; if(val > 0&&dis[v] == dis[u] + 1){ int f = dfs(v, t, min(maxflow, val)); if(f > 0) { edge[i].val -= f; edge[i ^ 1].val += f; return f; } } } return 0; } int dinic(int s,int t){ int flow = 0; while(bfs(s, t)){ while(int f = dfs(s, t, inf)) flow += f; } return flow; } int low[209][29],high[209][29],du[309]; char op[5]; int main(){ int T; scanf("%d", &T); while(T--){ memset(head, -1, sizeof(head)); memset(low, 0, sizeof(low)); memset(high, inf, sizeof(high)); memset(du, 0, sizeof(du)); gtot = 0; scanf("%d%d", &n, &m); int s = 0, t = n+m+1, ss = n+m+2, tt = n+m+3; all = tt; int s1 = 0, s2 = 0; for(int i=1; i<=n; i++) { int x; scanf("%d", &x); addedge(s, i, 0, -1); du[s] -= x; du[i] += x; s1 += x; } for(int i=1; i<=m; i++){ int x; scanf("%d", &x); addedge(n+i, t, 0, -1); du[t] += x; du[n+i] -= x; s2 += x; } int c; scanf("%d", &c); int flag = 1; while(c--){ int u,v,x; scanf("%d %d %s %d", &u, &v, op, &x); if(u == 0 && v == 0){ for(int i=1; i<=n; i++) { for(int j=1; j<=m; j++){ if(op[0] == ‘>‘) low[i][j] = max(low[i][j],x+1); else if(op[0] == ‘<‘) high[i][j] = min(high[i][j], x-1); else if(op[0] == ‘=‘) { low[i][j] = max(low[i][j], x), high[i][j] = min(high[i][j], x); if(low[i][j] != x || high[i][j] != x) flag = 0; } } } } else if(u == 0) { for(int i=1; i<=n; i++) { if(op[0] == ‘>‘) low[i][v] = max(low[i][v],x+1); else if(op[0] == ‘<‘) high[i][v] = min(high[i][v], x-1); else if(op[0] == ‘=‘) { low[i][v] = max(low[i][v], x), high[i][v] = min(high[i][v], x); if(low[i][v] != x || high[i][v] != x) flag = 0; } } } else if(v == 0){ for(int i=1; i<=m; i++) { if(op[0] == ‘>‘) low[u][i] = max(low[u][i],x+1); else if(op[0] == ‘<‘) high[u][i] = min(high[u][i], x-1); else { low[u][i] = max(low[u][i], x), high[u][i] = min(high[u][i], x); if(low[u][i] != x || high[u][i] != x) flag = 0; } } } else { if(op[0] == ‘>‘) low[u][v] = max(low[u][v],x+1); else if(op[0] == ‘<‘) high[u][v] = min(high[u][v], x-1); else { low[u][v] = max(low[u][v], x), high[u][v] = min(high[u][v], x); if(low[u][v] != x || high[u][v] != x) flag = 0; } } } for(int i=1; i<=n; i++) { for(int j=1; j<=m; j++){ du[i] -= low[i][j]; du[j+n] += low[i][j]; addedge(i, n + j, high[i][j] - low[i][j], 1); if(high[i][j] < low[i][j]) flag = 0; } } int sum = 0; for(int i=s; i<=t; i++) { if(du[i] > 0) addedge(ss, i, du[i], -1), sum += du[i]; if(du[i] < 0)addedge(i, tt, -du[i], -1); } if(s1 != s2 || !flag) { puts("IMPOSSIBLE"); if(T) puts(""); continue; } int f = dinic(ss, tt); if(f + s1==sum) { for(int i=n+1;i<=n+m; i++){ for(int j=head[i]; ~j; j=edge[j].nxt){ int v = edge[j].v,val = edge[j].val; low[v][i-n] += val; } } for(int i=1; i<=n; i++){ for(int j=1; j<=m; j++){ if(j < m)printf("%d ", low[i][j]); else printf("%d\n", low[i][j]); } } } else puts("IMPOSSIBLE"); if(T) puts(""); } return 0; } /* 2 2 3 8 10 5 6 7 4 0 2 > 2 2 1 = 3 2 3 > 2 2 3 < 5 2 2 4 5 6 7 1 1 1 > 10 */
poj2396 Budget 上下界可行流