标签：turn node memset 情况 long class efi image 输出
输入
5 7
1 1 2
2 3 2
4 4 1
3 4 5
5 4 5
2 3 5
4 5 1
输出
11
依然是分析建边：
\(X=1\)，有\(A=B\)，因为我们要建立不等关系，所以转化一下变成\(B>=A+0\)&&\(B>=A+0\)，这里因为要求最小值所以转化为>=跑最长路。
\(X=2\)，有\(A<B\)，转化得\(B>=A+1\)
\(X=3\)，有\(A>=B\)，转化得\(A>=B+0\)（不要吐槽这是废话qwq）
\(X=4\)，有\(A>B\)，转化得\(A>=B+1\)
\(X=5\)，有\(A<=B\)，转化得\(B>=A+0\)
因为要求分发的总量，搞一个超级源点向各点连边，边权为1，最后累加dis即可
PS:这道题有一个地方是第二条和第四条可以在读入时直接判掉一部分非法情况，因为同一个人糖果数不可能不同。如果不这样剪枝的会T掉两个点，当然你也可以选择最后0到各点加边时倒着来，至于为什么能行咱也不知道。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 100000 + 10;
#define ll long long
#define rint register int
int n, m, len = 0;
int head[maxn], dis[maxn], cnt[maxn];
bool vis[maxn];
struct edge{
int nex, to, w;
}e[maxn << 2];
void Add(int u, int v, int w){
e[++len].to = v;
e[len].nex = head[u];
e[len].w = w;
head[u] = len;
}
bool Spfa(int u){
queue<int> q;
for(int i = 1; i <= n; i++)
dis[i] = -1;
dis[u] = 0;
vis[u] = 1;
q.push(u);
while(!q.empty()){
int x = q.front();q.pop();
vis[x] = 0;
for(rint i = head[x]; i; i = e[i].nex){
int v = e[i].to;
if(dis[v] < dis[x] + e[i].w){
dis[v] = dis[x] + e[i].w;
if(!vis[v]){
if(++cnt[v] >= n) return 0;
vis[v] = 1;
q.push(v);
}
}
}
}
return 1;
}
int main(){
cin >> n >> m;
int x, u, v;
for(rint i = 1; i <= m; i++){
scanf("%d %d %d", &x, &u, &v);
if(x == 1) Add(u, v, 0), Add(v, u, 0);
else if(x == 2){
if(u == v){
cout << -1 << endl;
return 0;
}
Add(u, v, 1);
}
else if(x == 3) Add(v, u, 0);
else if(x == 4) {
if(u == v){
cout << -1 << endl;
return 0;
}
Add(v, u, 1);
}
else if(x == 5) Add(u, v, 0);
}
for(rint i = n; i >= 1; i--) Add(0, i, 1);//超级源点倒序加边，为什么会快呢
if(!Spfa(0)) printf("-1\n");
else{
ll ans = 0;
for(rint i = 1; i <= n; i++)
ans += dis[i];
cout << ans << endl;
}
return 0;
}
咱考试时的一道水题，明明第一遍敲对了最后还是脑抽改错了，简直。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
#define ll long long
const int maxn = 1000 + 10;
const int maxm = 5000 + 10;
struct node{
int x, y, dis;
}a[5000<<2], b[5000<<2];
int head[maxn], len = 0, dis[maxn];
int n, M, K;
struct edge{
int to, nex, w;
}e[maxm << 2];
void Add(int u, int v, int w){
e[++len].to = v;
e[len].nex = head[u];
e[len].w = w;
head[u] = len;
}
int cnt[maxn];
bool vis[maxn];
void Spfa(int u){
queue<int> q;
for(int i = 1; i <= n; i++)
dis[i] = 0x3f3f3f3f;
q.push(u);
vis[u] = 1;
dis[u] = 0;
while(!q.empty()){
int x = q.front(); q.pop();
vis[x] = 0;
for(int i = head[x]; i; i = e[i].nex){
int v = e[i].to;
if(dis[v] > dis[x] + e[i].w){
dis[v] = dis[x] + e[i].w;
if(!vis[v]){
cnt[v]++;
if(cnt[v] >= n){
dis[n] = -1;
return ;
}
vis[v] = 1;
q.push(v);
}
}
}
}
}
bool Cmp(node w, node p){
return w.x == p.x ? w.y < p.x : w.x < p.x;
}
int main(){
cin >> n >> M >> K;
/*for(int i = 1; i <= M; i++)
scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].dis);
for(int i = 1; i <= K; i++)
scanf("%d %d %d", &b[i].x, &b[i].y, &b[i].dis);
sort(a + 1, a + 1 + M, Cmp);
sort(b + 1, b + 1 + K, Cmp);*/
int u, v, w;
for(int i = 1; i <= M; i++){
scanf("%d %d %d", &u, &v, &w);
Add(u, v, w);
}
for(int i = 1; i <= K; i++){
scanf("%d %d %d", &u, &v, &w);
Add(v, u, -w);
}
for(int i = 1; i <= n; i++){
Add(0, i, 0);
}
Spfa(0);
if(dis[n] == -1){
printf("-1\n");
return 0;
}
Spfa(1);
if(dis[n] == 0x3f3f3f3f){
printf("-2\n");
return 0;
}
printf("%d\n", dis[n]);
return 0;
}
我已经懒得粘题了qwq，传送门：https://www.luogu.com.cn/problem/P2294
这道题要用到一点前缀和的思想qwq，将\(U->V=C\)理解为\(sum[V]-sum[U-1]>=C\) && \(sum[U-1]-sum[V]>=C\)，然后就是最长路板子了qwq
记得图不一定连通，最后记得每个点都要跑一下。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
#define ll long long
#define inf 9999997
const int maxn = 100000 + 10;
inline int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < ‘0‘ || ch > ‘9‘){if(ch == ‘-‘) w = -1; ch = getchar();}
while(ch >= ‘0‘ && ch <= ‘9‘) s = s * 10 + ch - ‘0‘, ch = getchar();
return s * w;
}
struct edge{
int nex, to, w;
}e[maxn << 2];
int n, m;
int head[maxn], len = 0;
int dis[maxn], cnt[maxn];
bool vis[maxn];
void Add(int u, int v, int w){
e[++len].to = v;
e[len].nex = head[u];
e[len].w = w;
head[u] = len;
}
bool Spfa(int s){
queue<int> q;
for(int i = 1; i <= n; i++) dis[i] = -inf;
q.push(s);
dis[s] = 0;
vis[s] = 1;
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = 0;
cnt[u]++;
if(cnt[u] >= n) return 0;
for(int i = head[u]; i; i = e[i].nex){
int v = e[i].to;
if(dis[v] < dis[u] + e[i].w){
dis[v] = dis[u] + e[i].w;
if(!vis[v]){
vis[v] = 1;
q.push(v);
}
}
}
}
return 1;
}
void Init(){
memset(head, 0, sizeof head);
memset(vis, 0, sizeof vis);
memset(cnt, 0, sizeof cnt);
len = 0;
}
int main(){
int t = read();
while(t--){
Init();
n = read(), m = read();
int u, v, w;
for(int i = 1; i <= m; i++){
u = read(), v = read(), w = read();
Add(v, u - 1, -w);
Add(u - 1, v, w);
}
int flag = 1;
for(int i = 0; i <= n; i++){
if(!cnt[i]){
if(!Spfa(i)){
flag = 0;
break;
}
}
}
if(flag) cout << "true" << endl;
else cout << "false" << endl;
}
return 0;
}
传送门：https://www.luogu.com.cn/problem/P1993
比前面的题都要裸的题qwq，直接看代码吧
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
#define ll long long
const int maxn = 10000 + 10;
inline int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < ‘0‘ || ch > ‘9‘){if(ch == ‘-‘) w = -1; ch = getchar();}
while(ch >= ‘0‘ && ch <= ‘9‘) s = s * 10 + ch - ‘0‘, ch = getchar();
return s * w;
}
struct edge{
int nex, to, w;
}e[maxn << 2];
int n, m;
int head[maxn], len = 0;
int dis[maxn], cnt[maxn];
bool vis[maxn];
void Add(int u, int v, int w){
e[++len].to = v;
e[len].nex = head[u];
e[len].w = w;
head[u] = len;
}
void Spfa(int u){
queue<int> q;
for(int i = 1; i <= n; i++)
dis[i] = 0x3f3f3f3f;
dis[u] = 0;
vis[u] = 1;
q.push(u);
while(!q.empty()){
int x = q.front();
q.pop();
vis[x] = 0;
for(int i = head[x]; i; i = e[i].nex){
int v = e[i].to;
if(dis[v] > dis[x] + e[i].w){
dis[v] = dis[x] + e[i].w;
if(!vis[v]){
if(++cnt[v] >= n){printf("No\n"); return ;}
vis[v] = 1;
q.push(v);
}
}
}
}
printf("Yes\n");
return ;
}
int main(){
n = read(), m = read();
int x, a, b, c;
for(int i = 1; i <= m; i++){
x = read();
if(x == 1){
a = read(); b = read(); c = read();
Add(a, b, -c);
}
else if(x == 2){
a = read(); b = read(); c = read();
Add(b, a, c);
}
else{
a = read(); b = read();
Add(a, b, 0), Add(b, a, 0);
}
}
for(int i = 1; i <= n; i++) Add(0 , i, 0);
Spfa(0);
return 0;
}
查分约束的几道简单例题(糖果，layout，狡猾的商人，小k的农场）
标签：turn node memset 情况 long class efi image 输出
原文地址：https://www.cnblogs.com/Zfio/p/13332732.html
